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).