Find V of the gas released by the interaction of 240 g of Mg with an excess of HCl solution if the gas yield is 65%.

First, we compose the equation for the substitution reaction and place the required coefficients in it: Mg + 2 HCl = MgCl2 + H2.
Amount of substance n = m / M, where m is the mass of the substance in grams, M is the molar mass in g / mol. By the condition, m magnesium = 240 g. We look at the molar mass (rounded to the nearest integer) Mg in the periodic table, it is equal to 24g / mol. Let’s find the amount of Mg substance.
n (Mg) = 240 g / 24 g / mol = 10 mol.
n (Mg) = n (H2), since the coefficients before Mg and H2 in the reaction equation are equal.
n (H2) = V / V molar, the molar volume is 22.4 l / mol. Let’s find the theoretical H2 output and take it as 100%.
V (theory) = 22.4 l / mol * 10 mol = 224 l.
According to the condition of the problem, the H2 output was 65% of the theoretically possible, we will compose the proportion, where x is the volume of the released gas: 224 / x = 100/65, hence x = 224 * 65/100
x = 145.6 l.
Answer 145.6 liters