Question

If the volume of H2 that is generated is measured to be 18.79 mL at STP and the reaction started with 0.032 g of Mg, what is the molar volume of the gas at STP in L/mole?

Answer 1

Mg + 2HCl   -------------------> MgCl2 + H2

volume of H2 = 18.79 mL = 0.01879 L

mass of Mg = 0.032 g

moles of Mg = mass / molar mass of Mg

                     = 0.032 / 24

                     = 0.00133

moles of Mg = 0.00133

moles of Mg = moles of H2

moles of H2 = 0.00133

molar volume of gas = volume of gas / moles of gas

                                  = 0.01879 / 0.00133

                                 = 14.13 L / mole

molar volume of gas =14.13 L / mole