Find the mass of carbon monoxide (IV) obtained by the action of hydrochloric acid on sodium carbonate weighing 110 g, containing impurities by mass fraction of 2%.
Let’s find the mass of Na2CO3 without impurities.
100% – 2% = 98%.
110 g – 100%,
x – 98%,
x = (110 g × 98%): 100% = 107.8 g.
Let’s find the amount of the substance Na2CO3.
n = m: M.
M (Na2CO3) = 106 g / mol.
n = 107.8 g: 106 g / mol = 1.02 mol.
Let’s find the quantitative ratios of substances.
Na2CO3 + 2HCl = 2NaCl + CO2 ↑ + H2O
There is 1 mol of CO2 for 1 mole of Na2CO3.
The substances are in quantitative ratios of 1: 1.
The amount of substance will be equal.
n (Na2CO3) = n (CO2) = 1.02 mol.
Let’s find the mass of CO2.
m = n × M.
M (CO2) = 44 g / mol.
m = 44 g / mol × 1.02 mol = 44.88 g.
Answer: 44.88 g.
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.