Final answer:
The mass of magnesium necessary to produce 40 ml of H2 at STP is calculated to be 0.0435 g, using the molar volume of a gas at STP and the stoichiometry of the relevant chemical reaction.
Step-by-step explanation:
To calculate the mass of magnesium necessary to evolve 40 ml of H2 at STP, we must first convert the volume of H2 to moles using the molar volume of a gas at STP (22.4 L/mol). Since the volume given is 40 ml (or 0.040 L), the moles of H2 can be calculated as:
0.040 L H2 x (1 mol H2 / 22.4 L) = 0.00179 mol H2 (at STP)
Using the stoichiometry of the reaction where 1 mol of Mg produces 1 mol of H2, the mass of Mg needed is:
0.00179 mol H2 x (24.305 g Mg / 1 mol Mg) = 0.0435 g Mg
Therefore, the mass of magnesium required is 0.0435 g.