When aluminum interacts with an excess of sodium hydroxide solution, a hydrogen volume of 4.48

When aluminum interacts with an excess of sodium hydroxide solution, a hydrogen volume of 4.48 was obtained. What mass of aluminum was spent on this process

Let’s find the amount of the substance 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.

n = V: Vn.

n = 4.48 L: 22.4 L / mol = 0.2 mol.

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

2Al + 2 NaOH + 6 H2O = 2 Na [Al (OH) 4] + 3H2 ↑.

For 3 mol of hydrogen there is 2 mol of aluminum nitrate.

Substances are in quantitative ratios 2: 3 = 1: 1.5.

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

n (Al) = 2 / 3n (H2) = 0.2: 1.5 = 0.13 mol.

2 mol Al – 3 mol H2

n mol Al – 0.2 mol H2.

n mol Al = (0.2 mol × 2 mol): 3 mol = 0.133 mol.

Let’s find the mass of aluminum.

m = n × M.

M (Al) = 27 g / mol.

m = 27 g / mol × 0.133 mol = 3.59 g.

Answer: 3.59 g.