Calculate the mass of aluminum chloride, which is formed by the interaction of 8.1 g of aluminum

Calculate the mass of aluminum chloride, which is formed by the interaction of 8.1 g of aluminum with chlorine. Determine the volume required for this reaction.

1.Let’s find the amount of aluminum substance by the formula:

n = m: M.

M (Al) = 27 g / mol.

n = 8.1 g: 27 g / mol = 0.3 mol.

2Al + 3Cl2 = 2AlCl3.

2. The amount of the substance aluminum and aluminum chloride are the same, since 2 mol of aluminum chloride accounts for 2 mol of aluminum. They are in ratios of 2: 2 or 1: 1.

n (AlCl3) = 0.3 mol.

M (AlCl3) = 27 + 35 × 3 = 132 g / mol.

m = n × M,

m = 0.3 mol × 132 g / mol = 39.6 g.

For 2 moles of aluminum, there are 3 moles of chlorine. The substances are in quantitative ratios of 2: 3, or 1: 1.5. The amount of chlorine will be 1.5 times greater than the amount of aluminum.

n (Cl2) = 0.3 mol × 1.5 = 0.45 mol.

3. Find the volume of chlorine.

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 = 22.4 L / mol × 0.45 mol = 10.08 L.

Answer: 1.08 L; 39.6 g