Question

Acetylene gas, c2h2, burns in oxygen to produce carbon dioxide and water. if 60.5 g co2 are produced when 22.6 g c2h2 react with sufficient oxygen, what is the percent yield for the reaction?

Answer 1

Answer: -

79.11%

Explanation: -

Actual mass of CO₂ = 60.5 g

Molar mass of CO₂ = 12 x 1 + 16 x 2 = 44 g / mol

Actual number of moles of CO₂ = $\frac{60.5 g}{44 g/ mol}$

= 1.375 mol of CO₂

Mass of C₂H₂ = 22.6 g

Molar mass of C₂H₂ = 12 x 2 + 1 x 2

= 26 g / mol

Number of moles of C₂H₂ = $\frac{22.6 g}{26 g/ mol}$

= 0.869 mol

The balanced chemical equation for the reaction is

2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O

From the equation we see that

2 mol of C₂H₂ gives 4 mol of CO₂

0.869 mol of C₂H₂ gives $\frac{4 mol CO₂}{2 mol C₂H₂ } 0.869 mol C₂H₂$

= 1.738 mol of CO₂

Percentage yield = $\frac{1.375 mol of CO₂{1.738 mol of CO₂}$ x 100

=79.11%