If the percent yield is 83.4%, what mass of oxygen is needed to produce 125g of HCN?

Question

Hydrogen cyanide gas is produced by reacting ammonia gas, oxygen gas, and methane gas according to the reaction below. If the percent yield is 83.4%, what mass of oxygen is needed to produce 125g of HCN?

2 NH₃ + 3 O2 + 2 CH₄ --> 2 HCN + 6 H₂O

1. What does the 125g of HCn represent?

1a. Can it be used in stoich? (yes or no)

1b. Can it be used in the % yield formula? (yes or no)

2. What is the first thing that needs to be calculated in this problem? (DO IT)

3. Now calculate the grams of O₂ required. (SHOW ALL WORK INCLUDING ANY CONVERSIONS)

Answer 1

2NH₃ + 3O₂ + 2CH₄ ==> 2HCN + 6H₂O .. balanced equation

(1). 125 g HCN (hydrogen cyanide) represents the mass of one of the products

a). It cannot be used in stoichiometry. Only moles (and sometimes volume) can be used.

b). Yes, it can be used in the % yield formula

(2). The first thing that we should calculate is the moles of HCN represented by 125 g of HCN. Another approach would be to account for the 83.4% yield, but that's a matter of choice. So, there are really 2 correct answer to this part (see below**). I'll ignore % yield until part 3.

molar mass HCN = 27.0 g / mol

125 g HCN x 1 mol HCN / 27.0 g = 4.63 mols HCN

(3). We can use dimensional analysis (unit factor) and mole ratios in the equation to determine mass (grams) of O₂:

molar mass O₂ = 32 g / mol

4.63 mols HCN x 3 mols O₂ / 2 mols HCN = 6.95 mols O₂ needed if 100% yield

Since yield is 83.4%, moles O2 needed = 6.95 mols / 0.834 = 8.33 mols O2 needed

Mass (grams) O₂ needed = 8.33 mols O₂ x 32 g / mol = 267 g O₂ needed

**For part (2), you could do this:

125 g / 0.834 = 150. g HCN (accounting for % yield)

150 g HCN x 1 mol / 27 g = 5.56 mols HCN

Then for part (3), you'd have..

5.56 mols HCN x 3 mols O₂ / 2 mol HCN x 32 g O₂ / mol = 267 g O₂