Final answer:
To produce 35 mL of hydrogen gas at STP, 0.0379 grams of magnesium are required based on the stoichiometry of the reaction Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g).
Step-by-step explanation:
The student is asking how many grams of magnesium (Mg) are required to produce 35 mL of hydrogen gas (H₂) at Standard Temperature and Pressure (STP) using magnesium and hydrochloric acid (HCl). The reaction in question is Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g).
First, we need to find the amount of H₂ in moles that corresponds to 35 mL. At STP (0°C, 1 atm), 1 mole of a gas occupies 22.4 L (22400 mL). Thus, the number of moles of H₂ can be calculated using the relation:
35 mL H₂ × (1 mole / 22400 mL) = 0.0015625 moles H₂.
Since the molar ratio of Mg to H₂ in the reaction is 1:1, the same number of moles of Mg is required. Next, we calculate the mass of Mg using its molar mass (24.305 g/mol):
0.0015625 moles Mg × 24.305 g/mol = 0.0379 g Mg.
Therefore, to produce 35 mL of H₂ at STP, 0.0379 grams of magnesium are needed.