An alloy of copper with tin weighing 13.26 g was treated with an excess of dilute sulfuric acid, and 1.792

An alloy of copper with tin weighing 13.26 g was treated with an excess of dilute sulfuric acid, and 1.792 liters of gas were released. Indicate the mass fraction (%) of copper in the alloy.

Given:
m alloy = 13.26 g
V (gas) = 1,792 l

To find:
ω (Cu) -?

Decision:
1) Write the reaction equations:
Cu + H2SO4 – the reaction does not proceed;
Sn + H2SO4 => SnSO4 + H2 ↑;
2) Calculate the amount of substance H2:
n (H2) = V / Vm = 1.792 / 22.4 = 0.08 mol;
3) Determine the amount of substance Sn:
n (Sn) = n (H2) = 0.08 mol;
4) Calculate the mass of Sn:
m (Sn) = n * M = 0.08 * 119 = 9.52 g;
5) Calculate the mass of Cu:
m (Cu) = m alloy – m (Sn) = 13.26 – 9.52 = 3.74 g;
6) Calculate the mass fraction of Cu:
ω (Cu) = m (Cu) * 100% / m alloy = 3.74 * 100% / 13.26 = 28.21%.

Answer: Mass fraction of Cu is 28.21%.