Question 4.16: Calculating the Quantity of a Substance Produced by Reaction......

Analyze

The two reactions given above are simultaneous reactions. Simultaneous reactions occur independently; thus, we have two conversion pathways to consider:

(1)  g alloy → g Al → mol Al \xrightarrow[]{(a)} mol  H_{2};

(2)  g alloy → g Mg → mol Mg \xrightarrow[]{(b)} mol  H_{2};

Pathways (1) and (2) are based on the first and second reactions, respectively. The total amount of H_{2} produced is obtained by adding together the amounts produced by each reaction. The conversion from mol Al to mol H_{2} requires a stoichiometric factor, labeled (a). The conversion from mol Mg to mol H_{2} requires a different stoichiometric factor, labeled (b).

Solve

Convert from g alloy to mol Al and from g alloy to mol Mg by using the mass percentages of Al and Mg and the molar masses of Al and Mg.          ? mol Al = 0.710 g alloy × \frac{70.0  g  Al}{100.0  g  alloy} × \frac{1   mol  Al}{26.98  g  Al}

= 0.0184 mol Al

? mol Mg = 0.710 g alloy × \frac{30.0  g  Mg}{100.0  g  alloy} × \frac{1  mol  Mg}{24.31  g  Mg}

= 8.76 × 10^{−3} mol Mg

Convert from mol Al to mol H_{2} and from mol Mg to mol H_{2} by using stoichiometric factors (a) and (b). The total number of moles of H_{2} is obtained by adding together the two independent contributions.          ? mol H_{2} = 0.0184  mol  Al × \underbrace{\frac{3  mol  H_{2}}{2  mol  Al} }_{(a)} + 8.76 × 10^{−3}   mol  Mg × \underbrace{\frac{1  mol  H_{2}}{1  mol  Mg} }_{(b)} = 0.0364  mol  H_{2}

Convert from mol H_{2}  to  g  H_{2} by using the molar mass of H_{2} as a conversion factor.          ? g H_{2} = 0.0364  mol  H_{2} × \frac{2.016  g  H_{2}}{1  mol  H_{2}} = 0.0734  g  H_{2}

An alternative approach is to combine the steps into a single line calculation, as shown below.

? g H_{2} = (0.710  g  alloy × \frac{70.0  g  Al}{100.0  g  alloy} × \frac{1   mol  Al}{26.98  g  Al} × \underbrace{\frac{3  mol  H_{2}}{2  mol  Al} }_{(a)} × \frac{2.016  g  H_{2}}{1  mol  H_{2}}) + (0.710  g  alloy × \frac{30.0  g  Mg}{100.0  g  alloy} × \frac{1  mol  Mg}{24.31  g   Mg} × \underbrace{\frac{1  mol  H_{2}}{1  mol  Mg} }_{(b)} × \frac{2.016  g  H_{2}}{1  mol  H_{2}}) = 0.0734  g  H_{2}

Assess

In this example, the composition of the alloy is given and we solved for the amount of H_{2} that is produced. The inverse problem, in which we are given the amount of H_{2} produced and are asked to determine the amounts of Al and Mg in the alloy, is a little harder to solve. See Practice Examples A and B below.