A mixture of sodium and sodium oxide weighing 6.5 g was treated with water, while 2.8 liters

A mixture of sodium and sodium oxide weighing 6.5 g was treated with water, while 2.8 liters of gas were released. Determine the mass fractions of the components in the mixture

Let’s find the amount of substance H2.

n = V: Vn

n = 2.8 L: 22.4 L / mol = 0.125 mol.

2 Na + 2 H2O = 2NaOH + H2 ↑.

There are 2 mol of Na per 1 mole of H2. The amount of Na substance is 2 times greater than the amount of H2 substance. Substances are in quantitative ratios of 2: 1.

n (Na) = 2n (H2) = 0.125 × 2 = 0.25 mol.

M (Na) = 23 g / mol.

m (Na) = 23 g / mol × 0.25 mol = 5.75 g.

Find the mass of Na2O in the mixture.

6.5 g – 5.75 g = 0.75 g.

w (Na) = (5.75 g: 6.5 g) × 100% = 88.46%.

w (Na2O) = (0.75 g: 6.5 g) × 100% = 11.54%.

Answer: 11.54%; 88.46%.