Determine the volume of methane that is formed by the interaction of aluminum carbide weighing 172.8 g with water.

Let’s find the amount of aluminum substance Al4C3.

M (Al4C3) = 144 g / mol.

n = m: M.

n (Al4C3) = 172.8 g: 144 g / mol = 1.2 mol.

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

Al4C3 + 12 HCl = 3CH4 + 4 AlCl3.

According to the reaction equation, there is 3 mol of CH4 for 1 mol of Al 4C3. The substances are in quantitative ratios of 1: 3. The amount of CH4 is 3 times more than the amount of Al4C3.

n (CH4) = 3n (Al4C3) = 1.2 × 3 = 3.6 mol.

Let’s find the volume of CH4.

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

V = 3.6 mol × 22.4 L / mol = 80.64 L.

Answer: 80.64 liters.

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