Question

Balanced Chemical Equation: 2H2(g)+O2(g)——> 2H2O(g)

How many grams of O2 are needed to produce 118 g of H2O if the reaction has a 78.2 % yield?

Answer 1

You must use 134 g O₂ to produce 118 g H₂O.

`\%\text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} * 100 \%`

`\text{Theoretical yield} = \text{Actual yield}* \frac{100\%}{\%\text{ yield}} = \text{118 g }* (100 \%)/(78.2 \%) = \text{150.9 g}`

M_r: 32.00 18.02

2H₂ + O₂ ⟶ 2H₂O

Moles of H₂O = 150.9 g H₂O × (1 mol H₂O/18.02 g H₂O) = 8.374 mol H₂O

Moles of O₂ = 8.374 mol H₂O × (1 mol O₂/2 mol H₂O) = 4.187 mol O₂

Mass of O₂ = 4.1877 mol O₂ × (32.00 g O₂/1 mol O₂) = 134 g O₂