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? asked in Science & MathematicsChemistry · 3 years ago

Mg metal reacts with HCl to produce hydrogen gas: Mg(s)+2HCl(a)+H2(g).how many grams of Mg are needed to prepare 5.00 L of H2 at 735mm Hg...?

And 18°C. Whoever can explain in steps the best. I will give ten points (Best answer).

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  • Roger the Mole
    Lv 7
    3 years ago
    Favorite Answer

    Write the balanced reaction equation:

    Mg(s) + 2 HCl(a) → H2(g) + MgCl2(a)

    Use the Ideal Gas formula to find the number of moles of H2:

    n = PV / RT = (735 mmHg) x (5.00 L) / ((62.36367 L mmHg/K mol) x (18 + 273) K) = 0.2025 mol H2

    Use the mole ratio to find the moles of Mg:

    (0.2025 mol H2) x (1 mol Mg / 1 mol H2) = 0.2025 mol Mg

    Use the molar mass of Mg to find the grams of Mg:

    (0.2025 mol Mg) x (24.30506 g Mg/mol) = 4.92 g Mg

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