The volume of gas released under the action of an excess of hydrochloric acid per 1.5 mol of calcium carbide is

CaC2 + 2HCl = CaCl2 + C2H2
based on the stoichiometry of the reaction, the number of moles of acetylene = the number of moles of calcium carbide = 1.5
V acetylene = 1.5 * 22.4 = 33.6 L
Rounding to the nearest 34 l

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