What volume, in liters, of H2 at STP is released when 8.66g of Mg reacts?

Mg metal reacts with HCl to produce hydrogen gas.

Mg(s) +2HCl(aq)-> MgCl2(aq) + H2(g)

3 Answers

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  • 7 years ago

    From the equation- 1 mol Mg will produce 1 mol H2

    Molar mass Mg = 24.30g/mol

    Moles Mg = 8.66/24.3 = 0.356 mol Mg

    This will produce 0.356 mol H2

    At STP 1 mol H2 has volume = 22.4L

    0.356 mol H2 at STP has volume = 0.356*22.4 = 7.98 L H2

  • 7 years ago

    8.66 g Mg / 24.3 g/mol = 0.356 mol Mg

    0.356 mol Mg X (1 mol H2 / 1 mol Mg) = 0.356 mol H2

    Volume H2 = 0.356 mol X 22.4 L/mol = 7.98 L H2

  • 7 years ago

    1 mole of Mg reacts with 2 moles of HCl. 24.31*8.66= 210.52 moles of Mg thus 2*210.52=421.04 moles of HCL. Volume= (n*R*T)/P= (421.04*8.3145*273.15)/100000=9.56 L

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