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The reaction is 2HCl(aq) + Mg(s) ==> MgCl2(aq) + H2(g)
3.143 moles HCl will yield 1/2 x 3.143 moles of H2 gas (see balanced equation stoichiometry)
Thus, 3.143 moles/2 = 1.5715 moles H2 gas will be produced
At STP 1 mole - 22.4 liters, thus 1.5715 moles x 22.4 L/mole = 35.20 liters of H2 gas formed

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The volume of hydrogen is 35,25 L.

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Q: What volume of hydrogen in L is produced from the complete reaction of 3.143 moles of hydrogen chloride with magnesium at STP Mg s 2 HCl aq MgCl2 aq H2 g?
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