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In the reaction Mg (s) + 2HCl (aq) H2 (g) + MgCl2 (aq), how many moles of hydrogen gas will be produced from 75.0 milliliters of a 1.0 M HCl in an excess of Mg?
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Start by finding the moles of HCl in 75.0 milliliters of a 1.0 M HCl. Then use a ratio of the moles of HCl and Hydrogen gas (\(H_2\)), and their coefficients. \(\sf \dfrac{moles~of HCl}{HCl's ~Coefficient}=\dfrac{moles~of~H_2}{H_2's~coefficient}\) Plug in moles of HCl and solve for moles of \(H_2\).
(0.0750 L) x (1.0 mol/L HCl) x (1 mol H2 / 2 mol HCl) = 0.0375 mol H2
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