A mixture of potassium hydroxide and potassium hydrogencarbonate was treated with an excess of hydrochloric acid.

A mixture of potassium hydroxide and potassium hydrogencarbonate was treated with an excess of hydrochloric acid. This formed 29.8 g of potassium chloride and released 2.24 liters of gas. Calculate the mass fraction (%) of potassium bicarbonate in the mixture.

KOH + HCl ex → KCl + H2O
KHCO3 + HCl ex → KCl + H2O + CO2 ↑
n (CO2) = 2.24 / 22.4 = 0.1 mol
m (KCl in the second reaction) = 0.1 x (39 + 35.5) = 7.45 g
m (KCl in the first reaction) = 29.8 – 7.45 = 22.35 g
n (KCl1) = 22.35 / 74.5 = 0.3 mol
m (mixtures KOH + KHCO3) = 0.1 x (39 + 1 + 12 + 48) + 0.3 x (39 + 1 + 16) = 0.1 x 100 + 0.3 x 56 = 10 + 16, 8 = 26.8 g
m (KHCO3) = 0.1 x 100 = 10 g
ω (KHCO3) = 10 / 26.8 x 100% = 37.3%
Answer: the mass fraction of bicarbonate in the mixture is 37.3%.

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