A mixture of iron powders with copper weighing 12 g was treated with a sulfate acid solution.

A mixture of iron powders with copper weighing 12 g was treated with a sulfate acid solution. Gas was released with a volume of 4.48 liters. What are the masses and quantities of iron and copper substances in the initial mixture?

Given:
m (mixture) = 12 g
V (gas) = 4.48 l

To find:
m (Cu) -?
m (Fe) -?
n (Cu) -?
n (Fe) -?

1) Cu + H2SO4 – the reaction will not proceed;
Fe + H2SO4 => FeSO4 + H2 ↑;
2) n (H2) = V (H2) / Vm = 4.48 / 22.4 = 0.2 mol;
3) n (Fe) = n (H2) = 0.2 mol;
4) m (Fe) = n (Fe) * M (Fe) = 0.2 * 56 = 11.2 g;
5) m (Cu) = m (mixture) – m (Fe) = 12 – 11.2 = 0.8 g;
6) n (Cu) = m (Cu) / M (Cu) = 0.8 / 64 = 0.0125 mol.

Answer: The mass of Fe is 11.2 g; Cu – 0.8 g; the amount of Fe substance is 0.2 mol; Cu – 0.0125 mol.