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Sulfuric acid dissolves aluminum metal according to the following reaction: 2Al(s)+3H2SO4(aq) −> Al2(SO4)3(aq)+3H2(g)
Suppose you wanted to dissolve an aluminum block with a mass of 14.9 g.
What minimum mass of H2SO4 would you need?
What mass of H2 gas would be produced by the complete reaction of the aluminum block?
Express your answer in grams.
♦ Relevant knowledge
The metal aluminum is the metals are able to react with acids and create the salt hydrogen gas develops in the course of the reaction. The chemical equation that is balanced during the reaction is utilized to calculate the mole-tomole ratio of reactants and products.
This problem can be solved using the stoichiometry.
Stoichiometry is the measurement of the number of moles in a chemical reaction that contains different reactants or products.
Stoichiometry, a branch in chemistry that studies quantitative relationships between elements and/or substances involved in chemical reactions, is one example.
This is how you calculate the number of moles in a substance:
[katex]n = \frac{m}{{\rm{M}}}[/katex]
Here, [katex]n[/katex]
(1) This is the balanced chemical reaction:
[katex]2{\rm{Al}}\left( s \right) + 3{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_4}\left( {aq} \right) \to {\rm{A}}{{\rm{l}}_{\rm{2}}}{\left( {{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)_{\rm{3}}}\left( {aq} \right) + 3{{\rm{H}}_2}\left( g \right)[/katex]
[katex]\begin{array}{c}\\{n_{{\rm{Al}}}} = \left( {\frac{{14.9{\rm{ g}}}}{{27{\rm{ g mo}}{{\rm{l}}^{ – 1}}}}} \right)\\\\ = 0.55{\rm{ mol}}\\\end{array}[/katex]
Convert this number into [katex]{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}[/katex] equivalent moles
[katex]{n_{{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}} = \left( {{n_{{\rm{Al}}}}} \right)\left( n \right)[/katex]
Here, [katex]n[/katex]
[katex]\begin{array}{c}\\{n_{{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}} = \left( {0.55{\rm{ mol}}} \right)\left( {\frac{3}{2}} \right)\\\\ = 0.825\;{\rm{mol}}\\\end{array}[/katex]
To calculate [katex]{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_4}[/katex]’s mass, you can reorder the equation (1)
[katex]m = n \cdot M[/katex]
[katex]\begin{array}{c}\\{m_{{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_4}}} = \left( {0.825{\rm{ mol}}} \right)\left( {98{\rm{ g mo}}{{\rm{l}}^{ – 1}}} \right)\\\\ = 80.85{\rm{ g}}\\\end{array}[/katex]
(2) This is the balanced chemical reaction:
[katex]2{\rm{Al}}\left( s \right) + 3{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_4}\left( {aq} \right) \to {\rm{A}}{{\rm{l}}_{\rm{2}}}{\left( {{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)_{\rm{3}}}\left( {aq} \right) + 3{{\rm{H}}_2}\left( g \right)[/katex]
[katex]\begin{array}{c}\\{n_{{\rm{Al}}}} = \frac{{14.9{\rm{ g}}}}{{27{\rm{ g mo}}{{\rm{l}}^{ – 1}}}}\\\\ = 0.55{\rm{ mol}}\\\end{array}[/katex]
Convert this number into equivalent moles for [katex]{{\rm{H}}_{\rm{2}}}[/katex]
[katex]{n_{{{\rm{H}}_{\rm{2}}}}} = \left( {{n_{{\rm{Al}}}}} \right)\left( n \right)[/katex]
Here, [katex]n[/katex]
[katex]\begin{array}{c}\\{n_{{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}} = \left( {0.55{\rm{ mol}}} \right)\left( {\frac{3}{2}} \right)\\\\ = 0.825\;{\rm{mol}}\\\end{array}[/katex]
To calculate the mass of [katex]{{\rm{H}}_{\rm{2}}}[/katex], you need to rewrite the equation (1)
[katex]m = n \cdot M[/katex]
mH2 = (0.825 mol)(2.0 gmol-1) = 1.65 g
Ans:
Part 1: Minimum mass [katex]{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_4}[/katex] required in the reaction is 80.85 g.
Part 2: The mass of [katex]{{\rm{H}}_{\rm{2}}}[/katex] produced in the reaction is 1.65 g.