% Yield and limiting reagent Homework(show work in a easy way for a highschooler to understand

Question

1. 2KClO3 → 2KCl + 3O2
2. If you start with 23.5 g of KClO3, how much KCl will you theoretically make?

1. In the lab you make 9.32 g of KCl. What is your percent yield?

1. Fe3O4 + 4 H2→ 3Fe + 4H2O
2. If you start with 0.122 g of H2, how much H2O do you theoretically make?

1. You make 0.745 g of H2O in the lab, what is your percent yield?

1. 2Al+ 3Cl2→ 2AlCl3

How many grams of AlCl3 could be produced from 34.0 g Al and 39.0 g Cl2?

1. N2 + 3H2 → 2NH3
2. How many grams of NH3 will be produced if you have 55.2 g N2 and 89.7 g of H2?

1. What is your limiting reagents?

1. C3H8 + 5O2 → 3CO2 + 4H2O
2. If you have 17.6 g of C3H8 and 36.2 g of O2, how much CO2 will be made?

1. What is your limiting reagent?

Answer 1

For the first problem: First calculate the molar mass of KClO₃:

K: 39.098

Cl: 35.446

O: 3x15.999 = 47.997

KClO₃: 122.541 g/mol

So 23.5 g = 23.5/122.541 = 0.19177 moles of KClO₃

Using the molar ratios, there are 2 moles of KClO₃ required for every 2 moles of 2KCl produced meaning that there are 0.19177 moles of KCl produced.

Then calculate the molar mass of KCl:

K: 39.098

Cl: 35.446

KCl: 74.544 g/mol

Therefore 0.19177 moles of KCl times 74.544 g/mol = 14.3 grams.of HCl

Since 9.32 g of KCl, the percent yield was 9.32/14.3 = 65.2%

The other problems are similar. For the problems regarding the limiting reactant, see which of the reactants has excess (more than what is needed) and then the other reactant will be the limiting reactant and can be used to calculate the results.