Question

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 14.9g .

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.

Answer 1

Concepts and reason

The concept used to solve this problem is based upon the stoichiometry.

Stoichiometry relates number of moles of various reactants and products present in a balanced chemical reaction.

Fundamentals

Stoichiometry is a branch of chemistry that deals with quantitative relationships between the elements and/or compounds involved in a chemical reaction.

The number of moles of a substance is calculated as follows:

$n = \frac{m}{{\rm{M}}}$ …… (1)

Here, $n$ is the number of moles, $m$ is the mass and $M$ is the molar mass of the substance.

(1)

The balanced chemical reaction is as follows:

$2{\rm{Al}}\left( s \right) + 3{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_4}\left( {aq} \right) \to {\rm{A}}{{\rm{l}}_{\rm{2}}}{\left( {{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)_{\rm{3}}}\left( {aq} \right) + 3{{\rm{H}}_2}\left( g \right)$

Substitute $14.9{\rm{ g}}$ for $m$ and $27{\rm{ g mo}}{{\rm{l}}^{ - 1}}$ for $M$ in the equation (1) to calculate the number of moles of aluminium.

$\begin{array}{c}\\{n_{{\rm{Al}}}} = \left( {\frac{{14.9{\rm{ g}}}}{{27{\rm{ g mo}}{{\rm{l}}^{ - 1}}}}} \right)\\\\ = 0.55{\rm{ mol}}\\\end{array}$

Then, convert this into equivalent moles of ${{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}$ needed by the expression as follows:

${n_{{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}} = \left( {{n_{{\rm{Al}}}}} \right)\left( n \right)$

Here, $n$ is the mole ratio, ${n_{{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}}$ is the number of moles of ${{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}$ and ${n_{{\rm{Al}}}}$ is the number of moles of aluminium.

Substitute $0.55{\rm{ mol}}$ for ${n_{{\rm{Al}}}}$ and $\frac{3}{2}$ for $n$ in the above equation.

$\begin{array}{c}\\{n_{{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}} = \left( {0.55{\rm{ mol}}} \right)\left( {\frac{3}{2}} \right)\\\\ = 0.825\;{\rm{mol}}\\\end{array}$

Rearrange the equation (1) to calculate the mass of ${{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_4}$ as follows:

$m = n \cdot M$

Substitute $0.825{\rm{ mol}}$ for $n$ and $98{\rm{ g mo}}{{\rm{l}}^{ - 1}}$ for $M$ in the above equation.

$\begin{array}{c}\\{m_{{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_4}}} = \left( {0.825{\rm{ mol}}} \right)\left( {98{\rm{ g mo}}{{\rm{l}}^{ - 1}}} \right)\\\\ = 80.85{\rm{ g}}\\\end{array}$

(2)

The balanced chemical reaction is as follows:

$2{\rm{Al}}\left( s \right) + 3{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_4}\left( {aq} \right) \to {\rm{A}}{{\rm{l}}_{\rm{2}}}{\left( {{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)_{\rm{3}}}\left( {aq} \right) + 3{{\rm{H}}_2}\left( g \right)$

Substitute $14.9{\rm{ g}}$ for $m$ and $27{\rm{ g mo}}{{\rm{l}}^{ - 1}}$ for $M$ in the equation (1) to calculate the number of moles of aluminium.

$\begin{array}{c}\\{n_{{\rm{Al}}}} = \frac{{14.9{\rm{ g}}}}{{27{\rm{ g mo}}{{\rm{l}}^{ - 1}}}}\\\\ = 0.55{\rm{ mol}}\\\end{array}$

Then, convert this into equivalent moles of ${{\rm{H}}_{\rm{2}}}$ gas needed by the expression as follows:

${n_{{{\rm{H}}_{\rm{2}}}}} = \left( {{n_{{\rm{Al}}}}} \right)\left( n \right)$

Here, $n$ is the mole ratio, ${n_{{{\rm{H}}_{\rm{2}}}}}$ is the number of moles of ${{\rm{H}}_{\rm{2}}}$ gas and ${n_{{\rm{Al}}}}$ is the number of moles of aluminium.

Substitute $0.55{\rm{ mol}}$ for ${n_{{\rm{Al}}}}$ and $\frac{3}{2}$ for $n$ in the above equation.

$\begin{array}{c}\\{n_{{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}} = \left( {0.55{\rm{ mol}}} \right)\left( {\frac{3}{2}} \right)\\\\ = 0.825\;{\rm{mol}}\\\end{array}$

Rearrange the equation (1) to calculate the mass of ${{\rm{H}}_{\rm{2}}}$ gas as follows:

$m = n \cdot M$

Substitute $0.825{\rm{ mol}}$ for $n$ and ${\rm{2}}{\rm{.0 g mo}}{{\rm{l}}^{ - 1}}$ for $M$ in the above equation.

$\begin{array}{c}\\{m_{{{\rm{H}}_{\rm{2}}}}} = \left( {0.825{\rm{ mol}}} \right)\left( {{\rm{2}}{\rm{.0 g mo}}{{\rm{l}}^{ - 1}}} \right)\\\\ = 1.65{\rm{ g}}\\\end{array}$

Ans: Part 1

The minimum mass of ${{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_4}$ required in the reaction is $80.85{\rm{ g}}$ .

Part 2

The mass of ${{\rm{H}}_{\rm{2}}}$ produced in the reaction is $1.65{\rm{ g}}$ .