Name _______KEY___________________________________ Date _________ Period ______
Unit 10: Stoichiometry Review
Balance the reaction below. Use it to answer problems 1-7.
__6__ H2O (g) + ____ Al2S3 (s) → __2__ Al(OH)3 (s) + __3__ H2S (g)
18.02
1.
150.17
34.09
If 3.12 moles of aluminum hydroxide are produced, how many moles of aluminum
sulfide reacted?
3.12 mol Al(OH)3 x
2.
78.01
1 mol Al2S3
= 1.56 mol Al2S3
2 mol Al(OH)3
If 14.7 L of H2S are produced, what volume of water vapor reacted?
6 L H2 O
= 29.4 L H2O
3 L H 2S
If you react 244.1 g of Al2S3, what volume of H2S can you expect?
14.7 L H2S x
3.
244.1 g Al2S3 x
4.
1 mol Al2S3
3 mol H2S
22.4 L H2S
x
x
= 109.2 L H2S
150.17 g Al2S3 1 mol Al2S3 1 mol H2S
How many grams of water react to produce 5.81 moles of aluminum hydroxide?
5.81 mol Al(OH)3 x
5.
How many moles of aluminum hydroxide will be produced with 25.63 g of H2S?
25.63 g H2S x
6.
1 mol H2S
2 mol Al(OH)3
x
= 0.5012 mol Al(OH)3
34.09 g H2S
3 mol H2S
How many grams of aluminum sulfide will react with 33.0 grams of water?
33.0 g H2O x
7.
6 mol H2O
18.02 g H2O
x
= 314 g H2O
2 mol Al(OH)3
1 mol H2O
1 mol H2O
1 mol Al2S3 150.17 g Al2S3
x
x
= 45.8 g Al2S3
18.02 g H2O
6 mol H2O
1 mol Al2S3
If 67.49 grams of aluminum sulfide react, how many grams of both H2S and Al(OH)3
were produced?
67.49 g Al2S3 x
1 mol Al2S3
2 mol Al(OH)3 78.01 g Al(OH)3
x
x
= 70.12 g Al(OH)3
150.17 g Al2S3
1 mol Al2S3
1 mol Al(OH)3
67.49 g Al2S3 x
1 mol Al2S3
3 mol H2S
34.09 g H2S
x
x
= 45.96 g H2S
150.17 g Al2S3
1 mol Al2S3
1 mol H2S