Reduction of 40 g of copper (II) oxide with carbon (II) oxide resulted in 28 g of copper.

Reduction of 40 g of copper (II) oxide with carbon (II) oxide resulted in 28 g of copper. What is the mass fraction of the reaction product yield?

CuO + CO = Cu + CO2

Let’s find the amount of copper oxide substance and the resulting metallic copper:

n (CuO) = m (CuO) / M (CuO) = 40/80 = 0.5 mol;

npractice (Cu) = mpractice (Cu) / M (Cu) = 28/64 = 0.4375 mol;

According to the stoichiometry of the reaction:

ntheor (Cu) = n (CuO) = 0.5 mol;

Let’s find the reaction output:

η = npract (Cu) / ntheor (Cu) = 0.4375 / 0.5 = 0.875 = 87.5%.

Answer: η = 87.5%.