Stoichiometry Stoichiometry: Calculations with Chemical Formulas and Equations.
Balancing Equations and Stoichiometry. Chemical Equations Terms: (s) = solid (l) = liquid (g) = gas ...
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Transcript of Balancing Equations and Stoichiometry. Chemical Equations Terms: (s) = solid (l) = liquid (g) = gas ...
![Page 1: Balancing Equations and Stoichiometry. Chemical Equations Terms: (s) = solid (l) = liquid (g) = gas = heat (aq) = aqueous solution.](https://reader036.fdocuments.in/reader036/viewer/2022080917/56649f035503460f94c174a5/html5/thumbnails/1.jpg)
Balancing Equations and Stoichiometry
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Chemical Equations
Terms:(s) = solid(l) = liquid(g) = gas
D = heat(aq) = aqueous
solution
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Balancing Equations• * Use coefficients to balance equations!
• Step 1: Balance metals first.
• Step 2: If possible, consider poly-atomic ions as a group. If “OH” is present on one side and H2O is present on the other side, break up water into H and OH.
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Balancing Equations
• Step 3: Balance other elements
Step 4: Balance H’s and O’s last.
• Step 4: Double-check.
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Sample Problem
• Balance the reaction:• Cu + AgNO3 Ag +
Cu(NO3)2
• Ca(OH)2 + H3PO4 H2O +
Ca3(PO4)2
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Stoichiometric Calculations
• Given the reaction:
• C3H8 + 5O2 CO2 +
4H2O
• Info: molar ratios
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Problem
• C3H8 + O2 CO2 +
4H2O
• If 25 grams of C3H8 is used, how much O2 is needed?
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Before anything else….
• I. Find the molar masses of each compound in the reaction.
• II. Calculate the mole of the compound given using the equation:
mole = gram / molar mass
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Solution
• 1. Balance equation.• 2. Get molar ratios from
balanced equation.• 3. Find actual moles using
given masses.
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Solution (cont.)
• 4. Re-adjust moles.• 5. Convert moles to grams
if required.
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Steps in Stoichiometry• 1. Get the molar masses of each cpd in
the equation.
• 2. If grams are given, convert grams to moles using the equation: mole = gram/molar mass
• 3. Balance the equation.
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• 4. If only 1 mass is given (Case I), there is no limiting reagent. Re-adjust each mole using the molar ratios from the balanced equation.
• 5. If more than 1 mass is given, there is a LIMITING REAGENT! Base all actual moles of needed reactant and desired product on the Limiting Reagent (not on the Excess)! (Case II)
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• 6. Convert all moles to grams using the equation: gram = mole x molar mass
• 7. Grams of Reactants must equal grams of products.
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Limiting and Excess Reagents
• Limiting reagent = limits the amt. of product that can form
• Excess Reagent = reagent that is over and above what is needed
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Case II Stoichiometry
• Has a limiting and excess reagent
• Case II applies when there are 2 or more given masses or moles
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• 5. Convert moles to grams, if needed.Gram = mole x molar
mass
• 6. Calculate % Yield and % Error, if needed.
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Determining the Limiting Reagent
• To determine the limiting reagent, divide all calculated moles by the coefficients in the balanced reaction. The smallest value is the Limiting Reagent.
• Please note: Do not use these values for the rest of your calculations. This is only for the IDENTIFICATION of the Limiting Reagent!
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Yields
• Theoretical Yield –the amount of product formed when the limiting reagent is totally consumed
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Yield
• Actual Yield - often given as percent yield% Yield = actual yield X 100
• theoretical yield