What is the mass fraction of sodium carbonate in a technical sample, if anything, when 24.94

What is the mass fraction of sodium carbonate in a technical sample, if anything, when 24.94 of this salt reacted with hydrochloric acid, 4.48 liters of carbon dioxide were released.

1.Let’s find the amount of substance carbon dioxide.

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.

2. We compose the reaction equation:

Na2CO3 + 2HCl = 2NaCl + H2O + CO2.

According to the reaction equation, 1 mol of sodium carbonate accounts for 1 mol of carbon dioxide. The substances are in quantitative ratios of 1: 1.

Then the amount of substance carbon dioxide and sodium carbonate will be equal.

n (Na2CO3) = 0.2 mol.

3.Let’s find how much sodium carbonate should be consumed according to theory (calculations) to get 0.2 mol of gas.

m (Na2CO3) = n × M.

M (Na2CO3) = 23 × 2 + 12 + 16 × 3 = 106 g / mol.

m (Na2CO3) = 106 g / mol × 0.2 mol = 21.2 g.

4. According to the condition of the problem, 24.94 g of sodium carbonate took part in the reaction.

Let’s find how much pure sodium carbonate will be in a technical sample.

24.94 – 100%,

21.2 – x%,

X = (21.2 x 100%): 24.94;

X = 85%.

100% – 85% = 15% (impurities).