0.9 Quantitative aspects of chemical change  (Page 9/10)

${\mathrm{C}}_3{\mathrm{H}}_8\left(\mathrm{g}\right)+5{\mathrm{O}}_2\left(\mathrm{g}\right)\to 3\mathrm{C}{\mathrm{O}}_2\left(\mathrm{g}\right)+4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

From the balanced equation, the ratio of oxygen to propane in the reactants is 5:1.

1 volume of propane needs 5 volumes of oxygen, therefore $2{\mathrm{dm}}^3$ of propane will need $10{\mathrm{dm}}^3$ of oxygen for the reaction to proceed to completion.

$\mathrm{Fe}\left(\mathrm{s}\right)+\mathrm{S}\left(\mathrm{s}\right)\to \mathrm{FeS}\left(\mathrm{s}\right)$

From the equation $1\mathrm{mole}$ of $\mathrm{Fe}$ gives $1\mathrm{mole}$ of $\mathrm{FeS}$ . Therefore, $\mathrm{0,1}\mathrm{moles}$ of iron in the reactants will give $\mathrm{0,1}\mathrm{moles}$ of iron sulphide in the product.

The mass of iron (II) sulphide that is produced during this reaction is $\mathrm{8,79}\mathrm{g}$ .

From the balanced equation, the mole ratio of $\mathrm{H}{}_2\mathrm{SO}{}_4$ in the reactants to $\left(\mathrm{NH}{}_4\right){}_2\mathrm{SO}{}_4$ in the product is 1:1. Therefore, $\mathrm{20,39}\mathrm{mols}$ of $\mathrm{H}{}_2\mathrm{SO}{}_4$ of $\left(\mathrm{NH}{}_4\right){}_2\mathrm{SO}{}_4$ .

The maximum mass of ammonium sulphate that can be produced is calculated as follows:

The maximum amount of ammonium sulphate that can be produced is $\mathrm{2,694}\mathrm{kg}$ .