Find the volume of hydrogen that is formed by the interaction of 108 grams of Al with HCl.

| February 25, 2021 | education

1.Find the amount of aluminum substance by the formula:

n = m: M.

M (Al) = 27 g / mol.

n = 108 g: 27 g / mol = 4 mol.

2. Let’s compose the reaction equation, find the quantitative ratios of substances.

2Al + 6HCl = 2AlCl3 + 3H2.

According to the reaction equation, there is 3 mol of hydrogen for 2 mol of aluminum. The substances are in quantitative ratios of 2: 3.

The amount of hydrogen will be 1.5 times more than the amount of aluminum.

n (H2) = 1.5n (Al) = 4 mol × 1.5 = 6 mol.

2 mol Al -3 mol H2,

4 mol Al – n mol H2,

n mol H2 = (4 mol × 3 mol): 2 mol = 6 mol.

3. Let’s find the volume of hydrogen.

V = Vn n, where Vn is the molar volume of gas equal to 22.4 l / mol, and n is the amount of substance.

V = 6 mol × 22.4 L / mol = 134.4 L.

Answer: 134.4 liters.



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