Question

3 g of copper and magnesium were added to the mixture of hydrochloric acid, and 1.12 L of hydrogen was released as a reaction. Calculate the mass fraction (%) of copper in this mixture

Answer 1

Cu₂Mg + 2HCl → MgCl₂ + H₂ + 2Cu

At STP 1 mole of gas occupies 22.4 L.

Proportion 1:

1 mol – 22.4 L

x – 1.12 L

$x = \frac{1 \times 1.12}{22.4} = 0.05 \;mol$

n(Cu₂Mg) = n(H₂) = 0.05 mol

MM(Cu₂Mg) = 151.39 g/mol

m(Cu₂Mg) $= 0.05 \times 151.39 = 7.57 \;g$

Proportion 2:

7.57 g – 100 %

3 g – y

$y = \frac{3 \times 100}{7.57} = 39.63$ %

Answer: 39.63 %