Technical Na with a mass of 50 g reacts with h2o and a gas with a volume of 22.4 liters

| May 17, 2021 | education

Technical Na with a mass of 50 g reacts with h2o and a gas with a volume of 22.4 liters is released. Find the mass fraction of impurities in Na.

Let’s write the reaction equation:

2Na + 2H2O = 2NaOH + H2 ↑

Let’s find the amount of substance of the released hydrogen gas:

v (H2) = V (H2) / Vm = 22.4 / 22.4 = 1 (mol).

According to the reaction equation, 1 mol of H2 is formed from 2 mol of Na, therefore:

v (Na) = v (H2) * 2 = 1 * 2 = 2 (mol).

Let’s find the mass of pure sodium in a sample of technical sodium:

m (Na) = v (Na) * M (Na) = 2 * 23 = 46 (g).

Let’s find the mass of impurities in a sample of technical sodium:

m (impurities) = m (Na) tech. – m (Na) = 50 – 46 = 4 (g).

Let’s find the mass fraction of impurities in a sample of technical sodium:

w (impurities) = (m (impurities) / m (Na) tech.) * 100 = (4/50) * 100 = 8 (%).

Answer: 8%.



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