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(B) The balanced chemical equation is: $2 \ Al(s) + 3 \ H_2SO_4(aq) \longrightarrow Al_2(SO_4)_3(aq) + 3 \ H_2(g)$Moles of $Al = \frac{5.4 \ g}{27.0 \

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$6.15$

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(B) The balanced chemical equation is: $2 \ Al(s) + 3 \ H_2SO_4(aq) \longrightarrow Al_2(SO_4)_3(aq) + 3 \ H_2(g)$Moles of $Al = \frac{5.4 \ g}{27.0 \ g \ mol^{-1}} = 0.2 \ mol$Moles of $H_2SO_4 = M 	imes V(L) = 5.0 \ mol \ L^{-1} 	imes 0.050 \ L = 0.25 \ mol$According to the stoichiometry,$2 \ mol$ of $Al$ requires $3 \ mol$ of $H_2SO_4$. For $0.2 \ mol$ of $Al$,required $H_2SO_4 = \frac{3}{2} 	imes 0.2 = 0.3 \ mol$.Since we have only $0.25 \ mol$ of $H_2SO_4$,$H_2SO_4$ is the limiting reagent.Moles of $H_2$ produced $= \frac{3}{3} 	imes 0.25 \ mol = 0.25 \ mol$.Using the ideal gas law $PV = nRT$:$V = \frac{nRT}{P} = \frac{0.25 \ mol 	imes 0.082 \ atm \ L \ mol^{-1} \ K^{-1} 	imes 300 \ K}{1.0 \ atm} = 6.15 \ L$.

Similar Questions

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