What mass of hydrochloric acid participated in the reaction if 11.2 liters of hydrogen

What mass of hydrochloric acid participated in the reaction if 11.2 liters of hydrogen were released during the interaction between acid and zinc?

To solve this problem, we write down the reaction equation: Zn + 2HCl = ZnCl2 + H2 We sign 11.2 liters over hydrogen, and over hydrochloric acid (HCl) = x gr. Calculate the molar mass of hydrochloric acid M (HCl) = 1 + 35.5 = 36.5 g / mol, multiply the molar mass by the number of acid atoms. which entered into a reaction and we get 73 gr. Under hydrogen, we write down a constant molar volume, which is 22.4 liters. Let’s make a calculation and find x. x = 73 g * 11.2 l / 22.4 l = 36.5 g.
Answer: the mass of hydrochloric acid that participated in the reaction is 36.5 g