What volume of carbon dioxide will be released when 15 g of sodium carbonate containing 10% impurities

| March 19, 2021 | education

What volume of carbon dioxide will be released when 15 g of sodium carbonate containing 10% impurities interacts with an excess of hydrochloric acid? calculate the mass of water that was formed by the interaction of 280 g of hydrochloric acid with magnesium hydroxide

Let’s find the volume of the evolved gas by the formula.

V = Vn n, where Vn is the molar volume of gas, equal to 22.4 l / mol, and n is the amount of substance.

1.Let’s find the mass of pure sodium carbonate without impurities.

100% – 10% = 90%.

15 kg – 100%,

X – 90%,

X = (15 kg × 90%): 100% = 13.5 g.

2.Let’s find the amount of sodium carbonate substance by the formula:

n = m: M.

M (Na2CO3) = 46 + 12 + 48 = 106 g / mol.

n = 13.5 g: 106 g / mol = 0.127 mol.

3. Let’s compose the equation of the reaction between sodium carbonate and hydrochloric acid. Let us determine in what quantitative ratios are carbon dioxide and sodium carbonate.

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

1 mole of sodium carbonate accounts for 1 mole of carbon dioxide. Substances are in quantitative ratios of 1: 1. The amount of salt and gas substances will be the same.

n (Na2CO3) = n (CO2) = 0.127 mol.

4. V = Vn n.

V (CO2) = 0.127 mol × 22.4 L / mol = 2.84 L.

Answer: 2.84 liters.



One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.