Question

A sample of 508.4 grams of copper completely reacted with oxygen to form 572.4 grams of a copper oxide product. how many grams of oxygen must have reacted?

Answer 1

According to law of conservation of mass, mass can neither be destroyed nor created in a chemical reaction. Thus, sum of masses of reactants must be equal to sum of masses of products in a reaction.

The chemical reaction is as follows:

$2Cu+O_{2}\rightarrow 2CuO$

Here, sum of masses of Cu and oxygen gas should be equal to CuO formed.

$2m_{Cu}+m_{O_{2}}=2m_{CuO}$

Thus, mass of oxygen will be:

$m_{O_{2}}=2(572.4-508.4)g=128 g$

This can be further proved as follows:

The balanced chemical reaction is as follows:

$2Cu+O_{2}\rightarrow 2CuO$

Here, 2 moles of Cu completely reacts with 1 mole of $O_{2}$ to give 2 moles of $CuO$.

Thus, 1 mole of Cu reacts with 0.5 moles of $O_{2}$ .

The mass of Cu is 508.4 and molar mass is 63.546 g/mol, number of moles can be calculated as follows:

$n=\frac{m}{M}=\frac{508.4 g}{63.546 g/mol}=8 mol$

Thus, number of moles of  $O_{2}$ reacting will be:

$n_{O_{2}}=8\times 0.5 mol=4 mol$

Molar mass of oxygen molecule is 32 g/mol thus, mass can be calculated as follows:

m=n×M=4 mol×32 g/mol=128 g/mol

This satisfies the law of conservation of mass.