Answer to Question #138718 in General Chemistry for Yara

Question

Given the following equation,

3 H2SO4 + 2 Al(OH)3 ® 6 H2O + Al2(SO4)3

Determine the percentage yield of the reaction above if 12.0 g of sulfuric acid reacts with enough aluminum hydroxide, and 3.20 g of H2O are actually produced.

Answer 1

12g sulfuric acid =12g/(98g/mol)

=0.122mole sulfuric acid

the following equation,

3 H2SO4 + 2 Al(OH)3 ® 6 H2O + Al2(SO4)3

From this reaction we can say that

3 mole sulfuric acid reacts and produced 6mole water.

So ,1.22 mole sulfuric acid reacts and produced =6 mole*0.122 mole/3 mole

=0.244 mole water

So,

0.244 mole water=(0.244mole*18g/mole)

=4.40g of water.

Actually 3.20g water produced.

So percentage of yield=3.20*100/4.40

=72.17%

Hence percentage of yield of the reaction is 72.17%