Two questions w/ percentage yield?

Question

1. For the reaction SO3+H20 produces H2SO4, calculate the percentage yield if 500. g. of sulfur trioxide react with excess water to produce 575g of sulfuric acid.

2. For the reaction 2Na+2H20 produces 2NaOH+H2, calculate the percentage yield if 80g of water react with excess sodium to produce 4.14g of hydrogen.


Please show your work so i can understand how to answer these questions. Thank U.

Answer 1

SO3 + H2O ---> H2SO4

Theoretical yield = 500 g SO3 * 1 mole SO3/80.0 g SO3 * 1 mole * 1 mole H2SO4/1 mole SO3 * 98.0 g H2SO4/mole = 613 g

% yield = 500 g / 613 g * 100% = 81.6%

2 Na + 2 H2O ----> 2 NaOH + H2

Theoretical yield = 80.0 g H2O * 1 mole/18 g H2O * 1 mole H2/2 moles H2O * 2.02 g H2/mole = 4.49 g H2

% yield = 4.14/4.49 * 100 = 92.2%