What theoretical volume of hydrogen gas should be produced in a chemical reaction of HCl(aq) with 0.0300 g of magnesium at 21.8 o C, when the dry gas pressure is 735.6 torr

The correct answer is 0.3088 L.

Explanation:

The reaction will be,

Mg + 2 HCl = MgCl2 + H2

The moles can be determined by using the formula,

Moles = Given mass/Molecular mass

Based on the given information, the weight of Mg is 0.0300 g, and the molecular mass of Mg is 24.30 g/mol.

Now putting the values we get,

Moles = 0.300 g/24.30 g/mol = 0.01235 mol

From the reaction, it can be seen that 1 mol of Mg produces 1 mol of H2 gas then 0.01235 mol of Mg will produce 0.1235 mol of H2.

The volume of hydrogen gas produced can be determined by using the gas law equation,

V = nRT/P

Here V is volume in L, P is the pressure in atm, R is the gas constant, which is 0.08206, and T is temperature in Kelvin, 21.8 degree C = 21.8 + 273.15 = 294.95 K.

P = 735.6 torr = 735.6 * 0.00132 atm = 0.9679 atm

V = 0.1235 mol * 0.08206 atm L/mol/K * 294.95 K/0.9679 atm

V = 0.3088 L

Hence, the volume of hydrogen gas produced is 0.3088 L.

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