Consider the following balanced chemical equation: 2Cu2O(s) + O2(g) → 4CuO(s) Determine the mass (in g) of CuO formed if 27.1 g of Cu2O reacts with 27.1 g of O2.

Answer 1

2Cu₂O(s) + O₂(g) ==> 4CuO(s) ... balanced equation

Find limiting reactant: there are several ways to do this, but I find the easiest is to find moles of each reactant and divide that value by the coefficient in the balanced equation. Thus we have...

For Cu₂O:

moles = 27.1 g x 143 g/mol = 0.1895 moles (÷2 = 0.095)

For O₂:

moles = 27.1 g x 1 mol/32 g = 0.847 moles (÷1 = 0.847)

LIMITING REACTANT = Cu₂O

Theoretical yield of CuO:

0.1895 moles Cu₂O x 4 moles CuO/2 mol Cu₂O x 79.5 g/mol = 7.53 g CuO