Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”....
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Transcript of Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”....
![Page 1: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/1.jpg)
Stoichiometry
![Page 2: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/2.jpg)
Stoichiometry comes from the Greek words for “element” and “measure”.
Stoichiometry takes information for one element/compound in a reaction and allows for the calculation of the amount of a different element/compound in the same reaction.
Obviously the place to start is…
![Page 3: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/3.jpg)
1. Predict and write a balanced chemical equation.
![Page 4: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/4.jpg)
1. Predict and write a balanced chemical equation.
2. Start a RR track and convert the given information to moles (if not already as moles).
![Page 5: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/5.jpg)
1. Predict and write a balanced chemical equation.
2. Start a RR track and convert the given information to moles (if not already as moles).
a. Use the molar mass (from the PT) to change grams → moles
![Page 6: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/6.jpg)
1. Predict and write a balanced chemical equation.
2. Start a RR track and convert the given information to moles (if not already as moles).
a. Use the molar mass (from the PT) to change grams → moles
b. Use 6.022 x 1023 to change atoms/molecules/formula units → moles
![Page 7: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/7.jpg)
1. Predict and write a balanced chemical equation.
2. Start a RR track and convert the given information to moles (if not already as moles).
a. Use the molar mass (from the PT) to change grams → moles
b. Use 6.022 x 1023 to change atoms/molecules/formula units → moles
c. Use 22.4 L (for gases only) to change L → moles
![Page 8: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/8.jpg)
1. Predict and write a balanced chemical equation.
2. Start a RR track and convert the given information to moles (if not already as moles).
a. Use the molar mass (from the PT) to change grams → moles
b. Use 6.022 x 1023 to change atoms/molecules/formula units → moles
c. Use 22.4 L (for gases only) to change L → moles
d. For water and dilute solutions, 1 mL = 1 g
![Page 9: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/9.jpg)
3. Convert the moles of the given to moles of the unknown using the mole ratio from the balanced chemical equation
a. Mole # of unknownMole # of known
![Page 10: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/10.jpg)
3. Convert the moles of the given to moles of the unknown using the mole ratio from the balanced chemical equation
a. Mole # of unknownMole # of known
4. Convert the moles of the unknown to the desired unit (see possible conversion factors in step 2)
![Page 11: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/11.jpg)
Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning
Step 1:
![Page 12: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/12.jpg)
Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning
Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2
![Page 13: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/13.jpg)
Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning
Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2
a. Find the moles of phosphoric acid
![Page 14: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/14.jpg)
Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning
Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2
a. Find the moles of phosphoric acid1.00 x 102 g K
![Page 15: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/15.jpg)
Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning
Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2
a. Find the moles of phosphoric acid1.00 x 102 g K 1 mol K
39.1 g K
![Page 16: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/16.jpg)
Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning
Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2
a. Find the moles of phosphoric acid1.00 x 102 g K 1 mol K 2 mol H3PO4
=39.1 g K 6 mol K
![Page 17: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/17.jpg)
Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning
Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2
a. Find the moles of phosphoric acid1.00 x 102 g K 1 mol K 2 mol H3PO4
= 0.853 mol H3PO439.1 g K 6 mol K
![Page 18: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/18.jpg)
Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning
Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2
a. Find the moles of phosphoric acid
b. Find the mass of phosphoric acid
1.00 x 102 g K 1 mol K 2 mol H3PO4= 0.853 mol H3PO4
39.1 g K 6 mol K
![Page 19: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/19.jpg)
Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning
Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2
a. Find the moles of phosphoric acid
b. Find the mass of phosphoric acid
1.00 x 102 g K 1 mol K 2 mol H3PO4= 0.853 mol H3PO4
39.1 g K 6 mol K
1.00 x 102 g K
![Page 20: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/20.jpg)
Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning
Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2
a. Find the moles of phosphoric acid
b. Find the mass of phosphoric acid
1.00 x 102 g K 1 mol K 2 mol H3PO4= 0.853 mol H3PO4
39.1 g K 6 mol K
1.00 x 102 g K 1 mol K
39.1 g K
![Page 21: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/21.jpg)
Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning
Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2
a. Find the moles of phosphoric acid
b. Find the mass of phosphoric acid
1.00 x 102 g K 1 mol K 2 mol H3PO4= 0.853 mol H3PO4
39.1 g K 6 mol K
1.00 x 102 g K 1 mol K 2 mol H3PO4
39.1 g K 6 mol K
![Page 22: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/22.jpg)
Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning
Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2
a. Find the moles of phosphoric acid
b. Find the mass of phosphoric acid
1.00 x 102 g K 1 mol K 2 mol H3PO4= 0.853 mol H3PO4
39.1 g K 6 mol K
1.00 x 102 g K 1 mol K 2 mol H3PO4 98.0 g H3PO4=
39.1 g K 6 mol K 1 mol H3PO4
![Page 23: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/23.jpg)
Example: 1.00 x 102 g of Potassium reacts with phosphoric acid, each time start from the beginning
Step 1: 2H3PO4 + 6K → 2K3PO4 + 3H2
a. Find the moles of phosphoric acid
b. Find the mass of phosphoric acid
1.00 x 102 g K 1 mol K 2 mol H3PO4= 0.853 mol H3PO4
39.1 g K 6 mol K
1.00 x 102 g K 1 mol K 2 mol H3PO4 98.0 g H3PO4= 83.5 g H3PO4
39.1 g K 6 mol K 1 mol H3PO4
![Page 24: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/24.jpg)
c. Find the formula units of phosphoric acid
![Page 25: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/25.jpg)
c. Find the formula units of phosphoric acid1.00 x 102 g K
![Page 26: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/26.jpg)
c. Find the formula units of phosphoric acid1.00 x 102 g K 1 mol K
39.1 g K
![Page 27: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/27.jpg)
c. Find the formula units of phosphoric acid1.00 x 102 g K 1 mol K 2 mol H3PO4
39.1 g K 6 mol K
![Page 28: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/28.jpg)
c. Find the formula units of phosphoric acid1.00 x 102 g K 1 mol K 2 mol H3PO4 6.022 x 1023 formula units H3PO4
=39.1 g K 6 mol K 1 mol H3PO4
![Page 29: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/29.jpg)
c. Find the formula units of phosphoric acid1.00 x 102 g K 1 mol K 2 mol H3PO4 6.022 x 1023 formula units H3PO4
= 5.13 x 1023 H3PO439.1 g K 6 mol K 1 mol H3PO4
![Page 30: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/30.jpg)
c. Find the formula units of phosphoric acid
d. Find the moles of potassium phosphate
1.00 x 102 g K 1 mol K 2 mol H3PO4 6.022 x 1023 formula units H3PO4= 5.13 x 1023 H3PO439.1 g K 6 mol K 1 mol H3PO4
![Page 31: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/31.jpg)
c. Find the formula units of phosphoric acid
d. Find the moles of potassium phosphate
1.00 x 102 g K 1 mol K 2 mol H3PO4 6.022 x 1023 formula units H3PO4= 5.13 x 1023 H3PO439.1 g K 6 mol K 1 mol H3PO4
1.00 x 102 g K
![Page 32: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/32.jpg)
c. Find the formula units of phosphoric acid
d. Find the moles of potassium phosphate
1.00 x 102 g K 1 mol K 2 mol H3PO4 6.022 x 1023 formula units H3PO4= 5.13 x 1023 H3PO439.1 g K 6 mol K 1 mol H3PO4
1.00 x 102 g K 1 mol K
39.1 g K
![Page 33: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/33.jpg)
c. Find the formula units of phosphoric acid
d. Find the moles of potassium phosphate
1.00 x 102 g K 1 mol K 2 mol H3PO4 6.022 x 1023 formula units H3PO4= 5.13 x 1023 H3PO439.1 g K 6 mol K 1 mol H3PO4
1.00 x 102 g K 1 mol K 2 mol K3PO4=
39.1 g K 6 mol K
![Page 34: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/34.jpg)
c. Find the formula units of phosphoric acid
d. Find the moles of potassium phosphate
1.00 x 102 g K 1 mol K 2 mol H3PO4 6.022 x 1023 formula units H3PO4= 5.13 x 1023 H3PO439.1 g K 6 mol K 1 mol H3PO4
1.00 x 102 g K 1 mol K 2 mol K3PO4= 0.853 mol K3PO4
39.1 g K 6 mol K
![Page 35: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/35.jpg)
c. Find the formula units of phosphoric acid
d. Find the moles of potassium phosphate
e. Find the mass of potassium phosphate
1.00 x 102 g K 1 mol K 2 mol H3PO4 6.022 x 1023 formula units H3PO4= 5.13 x 1023 H3PO439.1 g K 6 mol K 1 mol H3PO4
1.00 x 102 g K 1 mol K 2 mol K3PO4= 0.853 mol K3PO4
39.1 g K 6 mol K
![Page 36: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/36.jpg)
c. Find the formula units of phosphoric acid
d. Find the moles of potassium phosphate
e. Find the mass of potassium phosphate1.00 x 102 g K
1.00 x 102 g K 1 mol K 2 mol H3PO4 6.022 x 1023 formula units H3PO4= 5.13 x 1023 H3PO439.1 g K 6 mol K 1 mol H3PO4
1.00 x 102 g K 1 mol K 2 mol K3PO4= 0.853 mol K3PO4
39.1 g K 6 mol K
![Page 37: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/37.jpg)
c. Find the formula units of phosphoric acid
d. Find the moles of potassium phosphate
e. Find the mass of potassium phosphate1.00 x 102 g K 1 mol K
39.1 g K
1.00 x 102 g K 1 mol K 2 mol H3PO4 6.022 x 1023 formula units H3PO4= 5.13 x 1023 H3PO439.1 g K 6 mol K 1 mol H3PO4
1.00 x 102 g K 1 mol K 2 mol K3PO4= 0.853 mol K3PO4
39.1 g K 6 mol K
![Page 38: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/38.jpg)
c. Find the formula units of phosphoric acid
d. Find the moles of potassium phosphate
e. Find the mass of potassium phosphate1.00 x 102 g K 1 mol K 2 mol K3PO4
39.1 g K 6 mol K
1.00 x 102 g K 1 mol K 2 mol H3PO4 6.022 x 1023 formula units H3PO4= 5.13 x 1023 H3PO439.1 g K 6 mol K 1 mol H3PO4
1.00 x 102 g K 1 mol K 2 mol K3PO4= 0.853 mol K3PO4
39.1 g K 6 mol K
![Page 39: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/39.jpg)
c. Find the formula units of phosphoric acid
d. Find the moles of potassium phosphate
e. Find the mass of potassium phosphate1.00 x 102 g K 1 mol K 2 mol K3PO4 212 g K3PO4
=39.1 g K 6 mol K 1 mol K3PO4
1.00 x 102 g K 1 mol K 2 mol H3PO4 6.022 x 1023 formula units H3PO4= 5.13 x 1023 H3PO439.1 g K 6 mol K 1 mol H3PO4
1.00 x 102 g K 1 mol K 2 mol K3PO4= 0.853 mol K3PO4
39.1 g K 6 mol K
![Page 40: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/40.jpg)
c. Find the formula units of phosphoric acid
d. Find the moles of potassium phosphate
e. Find the mass of potassium phosphate1.00 x 102 g K 1 mol K 2 mol K3PO4 212 g K3PO4
= 181 g K3PO439.1 g K 6 mol K 1 mol K3PO4
1.00 x 102 g K 1 mol K 2 mol H3PO4 6.022 x 1023 formula units H3PO4= 5.13 x 1023 H3PO439.1 g K 6 mol K 1 mol H3PO4
1.00 x 102 g K 1 mol K 2 mol K3PO4= 0.853 mol K3PO4
39.1 g K 6 mol K
![Page 41: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/41.jpg)
f. Find the mass of the hydrogen
![Page 42: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/42.jpg)
f. Find the mass of the hydrogen1.00 x 102 g K
![Page 43: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/43.jpg)
f. Find the mass of the hydrogen1.00 x 102 g K 1 mol K
39.1 g K
![Page 44: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/44.jpg)
f. Find the mass of the hydrogen1.00 x 102 g K 1 mol K 3 mol H2
39.1 g K 6 mol K
![Page 45: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/45.jpg)
f. Find the mass of the hydrogen1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2
=39.1 g K 6 mol K 1 mol H2
![Page 46: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/46.jpg)
f. Find the mass of the hydrogen1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2
= 2.58 g H239.1 g K 6 mol K 1 mol H2
![Page 47: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/47.jpg)
f. Find the mass of the hydrogen
g. Find the liters of hydrogen
1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2= 2.58 g H2
39.1 g K 6 mol K 1 mol H2
![Page 48: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/48.jpg)
f. Find the mass of the hydrogen
g. Find the liters of hydrogen
1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2= 2.58 g H2
39.1 g K 6 mol K 1 mol H2
1.00 x 102 g K
![Page 49: Stoichiometry. Stoichiometry comes from the Greek words for “element” and “measure”. Stoichiometry takes information for one element/compound in a reaction.](https://reader035.fdocuments.in/reader035/viewer/2022070410/56649f265503460f94c3ca0e/html5/thumbnails/49.jpg)
f. Find the mass of the hydrogen
g. Find the liters of hydrogen
1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2= 2.58 g H2
39.1 g K 6 mol K 1 mol H2
1.00 x 102 g K 1 mol K
39.1 g K
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f. Find the mass of the hydrogen
g. Find the liters of hydrogen
1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2= 2.58 g H2
39.1 g K 6 mol K 1 mol H2
1.00 x 102 g K 1 mol K 3 mol H2
39.1 g K 6 mol K
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f. Find the mass of the hydrogen
g. Find the liters of hydrogen
1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2= 2.58 g H2
39.1 g K 6 mol K 1 mol H2
1.00 x 102 g K 1 mol K 3 mol H2 22.4 L H2=
39.1 g K 6 mol K 1 mol H2
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f. Find the mass of the hydrogen
g. Find the liters of hydrogen
1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2= 2.58 g H2
39.1 g K 6 mol K 1 mol H2
1.00 x 102 g K 1 mol K 3 mol H2 22.4 L H2= 28.6 L H2
39.1 g K 6 mol K 1 mol H2
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f. Find the mass of the hydrogen
g. Find the liters of hydrogen
h. Find the molecules of hydrogen
1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2= 2.58 g H2
39.1 g K 6 mol K 1 mol H2
1.00 x 102 g K 1 mol K 3 mol H2 22.4 L H2= 28.6 L H2
39.1 g K 6 mol K 1 mol H2
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f. Find the mass of the hydrogen
g. Find the liters of hydrogen
h. Find the molecules of hydrogen
1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2= 2.58 g H2
39.1 g K 6 mol K 1 mol H2
1.00 x 102 g K 1 mol K 3 mol H2 22.4 L H2= 28.6 L H2
39.1 g K 6 mol K 1 mol H2
1.00 x 102 g K
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f. Find the mass of the hydrogen
g. Find the liters of hydrogen
h. Find the molecules of hydrogen
1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2= 2.58 g H2
39.1 g K 6 mol K 1 mol H2
1.00 x 102 g K 1 mol K 3 mol H2 22.4 L H2= 28.6 L H2
39.1 g K 6 mol K 1 mol H2
1.00 x 102 g K 1 mol K
39.1 g K
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f. Find the mass of the hydrogen
g. Find the liters of hydrogen
h. Find the molecules of hydrogen
1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2= 2.58 g H2
39.1 g K 6 mol K 1 mol H2
1.00 x 102 g K 1 mol K 3 mol H2 22.4 L H2= 28.6 L H2
39.1 g K 6 mol K 1 mol H2
1.00 x 102 g K 1 mol K 3 mol H2
39.1 g K 6 mol K
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f. Find the mass of the hydrogen
g. Find the liters of hydrogen
h. Find the molecules of hydrogen
1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2= 2.58 g H2
39.1 g K 6 mol K 1 mol H2
1.00 x 102 g K 1 mol K 3 mol H2 22.4 L H2= 28.6 L H2
39.1 g K 6 mol K 1 mol H2
1.00 x 102 g K 1 mol K 3 mol H2 6.022 x 1023 molecules H2=
39.1 g K 6 mol K 1 mol H2
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f. Find the mass of the hydrogen
g. Find the liters of hydrogen
h. Find the molecules of hydrogen
1.00 x 102 g K 1 mol K 3 mol H2 2.02 g H2= 2.58 g H2
39.1 g K 6 mol K 1 mol H2
1.00 x 102 g K 1 mol K 3 mol H2 22.4 L H2= 28.6 L H2
39.1 g K 6 mol K 1 mol H2
1.00 x 102 g K 1 mol K 3 mol H2 6.022 x 1023 molecules H2= 7.70 x 1023 molecules H239.1 g K 6 mol K 1 mol H2
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Limiting Reactant
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Rarely in real life does a reaction start with exactly proportional amounts of the reactants, as the smallest differences in mass still add up to more than trillions of atoms, and often there are big differences in the amounts of the reactants available.
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To determine which reactant will limit the reaction, the initial amount of all reactants must be known. Then take each react in turn and solve for the same product (it will not matter which product, so choose something convenient). The reactant the can produce the smallest amount of products is the limiting reactant, and all the others are the excess reactants.
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Example: 10.0 grams of sodium is added to 50.0 grams of water. Which reactant is the limiting reactant and which one is the excess reactant?
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Example: 10.0 grams of sodium is added to 50.0 grams of water. Which reactant is the limiting reactant and which one is the excess reactant?
Step 1: 2Na + 2HOH → 2NaOH + H2
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Step 2: Use both reactants to solve for the same product (let’s choose NaOH).
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Step 2: Use both reactants to solve for the same product (let’s choose NaOH).
10.0 g Na
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Step 2: Use both reactants to solve for the same product (let’s choose NaOH).
10.0 g Na 1 mol Na
23.0 g Na
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Step 2: Use both reactants to solve for the same product (let’s choose NaOH).
10.0 g Na 1 mol Na 2 mol NaOH
23.0 g Na 2 mol Na
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Step 2: Use both reactants to solve for the same product (let’s choose NaOH).
10.0 g Na 1 mol Na 2 mol NaOH 40.0 g NaOH=
23.0 g Na 2 mol Na 1 mol NaOH
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Step 2: Use both reactants to solve for the same product (let’s choose NaOH).
10.0 g Na 1 mol Na 2 mol NaOH 40.0 g NaOH= 17.4 g NaOH
23.0 g Na 2 mol Na 1 mol NaOH
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Step 2: Use both reactants to solve for the same product (let’s choose NaOH).
10.0 g Na 1 mol Na 2 mol NaOH 40.0 g NaOH= 17.4 g NaOH
23.0 g Na 2 mol Na 1 mol NaOH
50.0 g H2O
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Step 2: Use both reactants to solve for the same product (let’s choose NaOH).
10.0 g Na 1 mol Na 2 mol NaOH 40.0 g NaOH= 17.4 g NaOH
23.0 g Na 2 mol Na 1 mol NaOH
50.0 g H2O 1 mol H2O
18.0 g H2O
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Step 2: Use both reactants to solve for the same product (let’s choose NaOH).
10.0 g Na 1 mol Na 2 mol NaOH 40.0 g NaOH= 17.4 g NaOH
23.0 g Na 2 mol Na 1 mol NaOH
50.0 g H2O 1 mol H2O 2 mol NaOH
18.0 g H2O 2 mol H2O
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Step 2: Use both reactants to solve for the same product (let’s choose NaOH).
10.0 g Na 1 mol Na 2 mol NaOH 40.0 g NaOH= 17.4 g NaOH
23.0 g Na 2 mol Na 1 mol NaOH
50.0 g H2O 1 mol H2O 2 mol NaOH 40.0 g NaOH=
18.0 g H2O 2 mol H2O 1 mol NaOH
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Step 2: Use both reactants to solve for the same product (let’s choose NaOH).
10.0 g Na 1 mol Na 2 mol NaOH 40.0 g NaOH= 17.4 g NaOH
23.0 g Na 2 mol Na 1 mol NaOH
50.0 g H2O 1 mol H2O 2 mol NaOH 40.0 g NaOH= 111 g NaOH
18.0 g H2O 2 mol H2O 1 mol NaOH
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Step 2: Use both reactants to solve for the same product (let’s choose NaOH).
Na is limiting and thus H2O is excess.
10.0 g Na 1 mol Na 2 mol NaOH 40.0 g NaOH= 17.4 g NaOH
23.0 g Na 2 mol Na 1 mol NaOH
50.0 g H2O 1 mol H2O 2 mol NaOH 40.0 g NaOH= 111 g NaOH
18.0 g H2O 2 mol H2O 1 mol NaOH
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How much excess is there?Step 1: Starting with the limiting...
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How much excess is there?Step 1: Starting with the limiting...
10.0 g Na
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How much excess is there?Step 1: Starting with the limiting...
10.0 g Na 1 mol Na
23.0 g Na
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How much excess is there?Step 1: Starting with the limiting...
10.0 g Na 1 mol Na 2 mol H2O
23.0 g Na 2 mol Na
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How much excess is there?Step 1: Starting with the limiting...
10.0 g Na 1 mol Na 2 mol H2O 18.0 g H2O=
23.0 g Na 2 mol Na 1 mol H2O
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How much excess is there?Step 1: Starting with the limiting...
10.0 g Na 1 mol Na 2 mol H2O 18.0 g H2O= 7.83 g H2O used
23.0 g Na 2 mol Na 1 mol H2O
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How much excess is there?Step 1: Starting with the limiting...
Step 2: 50.0 g H2O – 7.83 g H2O = 42.17 g H2O
10.0 g Na 1 mol Na 2 mol H2O 18.0 g H2O= 7.83 g H2O used
23.0 g Na 2 mol Na 1 mol H2O
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How many grams of base are produce?
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How many grams of base are produce?17.4 g NaOH
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How many grams of base are produce?17.4 g NaOH
How many milliliters of gas is produced?
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How many grams of base are produce?17.4 g NaOH
How many milliliters of gas is produced?10.0 g Na
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How many grams of base are produce?17.4 g NaOH
How many milliliters of gas is produced?10.0 g Na 1 mol Na
23.0 g Na
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How many grams of base are produce?17.4 g NaOH
How many milliliters of gas is produced?10.0 g Na 1 mol Na 1 mol H2
23.0 g Na 2 mol Na
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How many grams of base are produce?17.4 g NaOH
How many milliliters of gas is produced?10.0 g Na 1 mol Na 1 mol H2 22. 4 L
23.0 g Na 2 mol Na 1 mol H2 gas
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How many grams of base are produce?17.4 g NaOH
How many milliliters of gas is produced?10.0 g Na 1 mol Na 1 mol H2 22. 4 L 1000 mL
= 23.0 g Na 2 mol Na 1 mol H2 gas 1 L
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How many grams of base are produce?17.4 g NaOH
How many milliliters of gas is produced?10.0 g Na 1 mol Na 1 mol H2 22. 4 L 1000 mL
= 4870 mL23.0 g Na 2 mol Na 1 mol H2 gas 1 L