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

March 27, 2021 | education

| **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

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