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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answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :