Stoichiometry Practice Problem
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Transcript of Stoichiometry Practice Problem
Question
How many grams of water are produced when 5.14 grams of Hydrogen Nitrite is reacted with Barium Hydroxide?
Convert the Given Amount Into Moles
2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2
5.14 g1.0079 g
+14.007 g
+2(15.999) g
47.013 g
5.14 g x 1 mole
1 47.013 g
.109 moles
Find Moles for Each of the Other Chemicals
2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2
5.14 g Moles? Moles? Moles?
1.0079 g
+14.007 g
+2(15.999) g
47.013 g
5.14 g x 1 mole
1 47.013 g
.109 moles
In Each of the Columns Write the Moles of Given (x) a Fraction
2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2
5.14 g .109 moles x ?/? = .109 moles x ?/? = .109 moles x ?/? =
1.0079 g
+14.007 g
+2(15.999) g
47.013 g
5.14 g x 1 mole
1 47.013 g
.109 moles
The Numerator of the Fraction is the Coefficient of the That Column
2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2
5.14 g .109 moles x 1/? = .109 moles x 2/? = .109 moles x 1/? =
1.0079 g
+14.007 g
+2(15.999) g
47.013 g
5.14 g x 1 mole
1 47.013 g
.109 moles
The Denominator of the Fraction is the Coefficient of the Given Column
2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2
5.14 g .109 moles x ½ = .109 moles x 2/2 = .109 moles x ½ =
1.0079 g
+14.007 g
+2(15.999) g
47.013 g
5.14 g x 1 mole
1 47.013 g
.109 moles
Do Math and Label as Moles
2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2
5.14 g .109 moles x ½ = .109 moles x 1 = .109 moles x ½ =
1.0079 g
+14.007 g
+2(15.999) g
47.013 g
.0545 moles .109 moles .0545 moles
5.14 g x 1 mole
1 47.013 g
.109 moles
Convert All Moles Into Grams
2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2
5.14 g .109 moles x ½ = .109 moles x 1 = .109 moles x ½ =
1.0079 g
+14.007 g
+2(15.999) g
47.013 g
.0545 moles .109 moles .0545 moles137.33 g
+ 2(15.999) g
+ 2(1.0079) g
171.34 g
2(1.0079) g
+ 15.999 g
18.015 g
137.3 g
+ 2(14.007) g
+ 4(15.999) g
229.34 g5.14 g x 1 mole
1 47.013 g .0545 moles x 171.34 g
1 1 mole
.109 moles x 18.015 g
1 1 mole
.0545 moles x 229.34 g
1 1 mole
.109 moles 9.34 g 1.96 g 12.5 g
Verify the Law of Conservation and Mass
2H(NO2) + Ba(OH)2 2H2O + Ba(NO2)2
5.14 g .109 moles x ½ = .109 moles x 1 = .109 moles x ½ =
1.0079 g
+14.007 g
+2(15.999) g
47.013 g
.0545 moles .109 moles .0545 moles137.33 g
+ 2(15.999) g
+ 2(1.0079) g
171.34 g
2(1.0079) g
+ 15.999 g
18.015 g
137.3 g
+ 2(14.007) g
+ 4(15.999) g
229.34 g5.14 g x 1 mole
1 47.013 g .0545 moles x 171.34 g
1 1 mole
.109 moles x 18.015 g
1 1 mole
.0545 moles x 229.34 g
1 1 mole
.109 moles 9.34 g 1.96 g 12.5 g
14.48 g 14.46 g