A mixture of aluminum and copper weighing 54 g was treated with hydrochloric acid. Collected 33.6 liters of gas.

A mixture of aluminum and copper weighing 54 g was treated with hydrochloric acid. Collected 33.6 liters of gas. Mass fraction of copper in the mixture?

To solve this problem, we write down the given: m (mixture) = 54 g, as a result of interaction with hydrochloric acid (HCl), a gas with a volume of 33.6 liters was formed
Find: w (Cu) -?
Decision:
Let’s write down the reaction equations.
Al + HCl = AlCl3 + H2
Copper will not interact with hydrochloric acid, since it is among the activities of metals after hydrogen.
Let’s arrange the coefficients.
2Al + 6HCl = 2AlCl3 + 3H2
Let’s calculate the mass of aluminum that has reacted.
Over hydrogen we write 33.6 l, and under hydrogen we write a constant molar volume, which is 22.4 l / mol.
Since 3 mol of hydrogen was formed, 3 mol * 22.4 l / mol = 67.2 l
Above aluminum we write x g, and under aluminum its molar mass, which is 27 g / mol. Since aluminum entered 2 mol, you need 2 mol * 27 g / mol = 54 g
Let’s compose and solve the proportion:
x = 54 * 33.6 / 67.2 = 27 g
Let’s calculate the mass of copper that is included in the mixture.
m (Cu) = 54 – 27 = 27 g
Let’s write down the formula for calculating the mass fraction:
w = m (substances) / m (mixtures)
w (Cu) = 27/54 = 0.5 or 50%
Answer: w (Cu) = 0.5 or 50%