Stoichiometry of metal–acid reaction: Six grams of magnesium (atomic weight = 24) react completely with excess acid. What mass of hydrogen gas (H2) is produced (in grams)?

Introduction / Context:Metal–acid reactions liberate hydrogen gas according to the general pattern M + 2H⁺ → M²⁺ + H₂ (for a divalent metal). This problem is a straightforward stoichiometry exercise using molar masses and reaction coefficients to determine the mass of hydrogen evolved.

Given Data / Assumptions:

• Metal: magnesium, atomic weight = 24 g/mol.
• Acid present in excess; magnesium is the limiting reagent.
• Reaction: Mg + 2H⁺ → Mg²⁺ + H₂ (1:1 molar ratio between Mg and H₂).

Concept / Approach:Compute moles of Mg from its mass and molar mass. Use the stoichiometric 1:1 ratio to obtain moles of H₂. Multiply by the molar mass of H₂ (2 g/mol) to obtain mass of hydrogen produced.

Step-by-Step Solution:n_Mg = 6 g / 24 g·mol^-1 = 0.25 mol.Stoichiometry: 1 mol Mg → 1 mol H₂, so n_H2 = 0.25 mol.Mass of H₂ = n_H2 * M_H2 = 0.25 mol * 2 g·mol^-1 = 0.5 g.Therefore, 0.5 g of hydrogen is produced.

Verification / Alternative check:Using moles conserves atoms: magnesium atoms transfer two electrons to two protons, pairing to form one molecule of H₂ per Mg atom, confirming the 1:1 mole ratio.

Why Other Options Are Wrong:1 g or 3 g and 5 g do not match the stoichiometric calculation with correct molar masses.

Common Pitfalls:

• Using a 2:1 ratio mistakenly (confusing with reaction coefficients).
• Using 1 g/mol for H rather than 2 g/mol for H₂ when converting to mass.

Final Answer:0.5