What volume of hydrogen is formed by the interaction of hydrochloric acid with 540 mg

What volume of hydrogen is formed by the interaction of hydrochloric acid with 540 mg of aluminum containing 4% impurities? How much of the salt substance will you get?

The reaction of aluminum with hydrochloric acid is described by the following chemical reaction equation.

Al + 3HCl = AlCl3 + 1.5 H2;

When 1 mol of aluminum is dissolved, 1.5 mol of hydrogen is released. This reaction will require 3 moles of hydrogen chloride.

To find the mass of pure aluminum, multiply the weight of the sample by the mass fraction of aluminum.

The mass of aluminum in the sample will be 0.54 x 0.96 = 0.518 grams;

Let’s find the amount of the substance contained in 0.518 grams of aluminum.

M Al = 27 grams / mol;

N Al = 0.518 / 27 = 0.019 mol;

Reaction with 0.019 mol of aluminum will result in 0.019 x 1.5 = 0.029 mol of hydrogen gas.

This forms 0.019 mol of aluminum chloride.

One mole of ideal gas under normal conditions takes a volume of 22.4 liters.

Determine the volume of hydrogen gas.

V H2 = 0.029 x 22.4 = 0.65 liters;