# 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%. Remember: The process of learning a person lasts a lifetime. The value of the same knowledge for different people may be different, it is determined by their individual characteristics and needs. Therefore, knowledge is always needed at any age and position.