9.4.10 Calculations Based on Chemical Equations
This is the eighth lecture from Chapter 4: “Stoichiometry” in the new Class 9 Chemistry book (Punjab Board – PCTB). It covers the chemical calculations based on chemical equations.
Example 1
25g of limestone (CaCO3) reacts with an excess of hydrochloric acid according to the above equation. How much calcium chloride (CaCl2) will be produced?
$$\begin{aligned}
\text{Mass of CaCO}_3 &= 25\ \text{g} \\
\text{Molar mass of CaCO}_3 &= 100\ \text{g mol}^{-1} \\
\text{Mass of CaCl}_2 &= ? \\
\text{Molar mass of CaCl}_2 &= 111\ \text{g mol}^{-1} \\
\\
100\ \text{g of CaCO}_3 &= 111\ \text{g of CaCl}_2 \\
1\ \text{g of CaCO}_3 &= \frac{111}{100}\ \text{g of CaCl}_2 \\
25\ \text{g of CaCO}_3 &= \frac{111}{100} \times 25 = 27.75\ \text{g of CaCl}_2
\end{aligned}$$
Example 2
1.80 moles of ethyl alcohol, when burnt in air completely, will utilize how many moles of oxygen gas? Also calculate the number of moles of CO2 produced.
$$ \displaylines{ \text{The balanced chemical equation for the reaction} \cr \mathrm{C_2H_5OH}_{(l)} \ + \ \mathrm{3O_2}_{(g)} \ \longrightarrow \ \mathrm{2CO_2}_{(g)} \ + \ \mathrm{3H_2O}_{(g)} \cr \cr \begin{aligned} \text{Moles of ethyl alcohol} &= 1.80 \text{moles} \\ \text{Moles of oxygen needed} &= \ ? \\ \\ \text{1 mole of ethyl alcohol needs oxygen } &= \text{ 3 moles}\\ \text{1.8 moles of ethyl alcohol will need oxygen } &= \frac{3}{1} \ \times \ 1.8 \\ &= \text{ 5.4 moles}\\ \\ \text{1 mole of ethyl alcohol produces CO}_2 &= \text{ 2.0 moles}\\ \text{1.8 moles of ethyl alcohol will produce CO}_2 &= \frac{2}{1} \ \times \ 1.8 \\ &= \text{ 3.6 moles}\\ \end{aligned}}$$Example 3
$$\displaylines{
\text{The balanced chemical equation for the reaction} \cr
\mathrm{4Al}_{\ (s)} \ + \ \mathrm{3O_2}_{\ (g)} \ \longrightarrow \ \mathrm{2Al_2O_3}_{\ (s)} \cr
\begin{aligned}
\text{Moles of Al } &= \ 0.3 \cr
\text{Grams of O}_2 \ \text{used } &= \ ? \cr
\cr
\text{4 moles of Al need oxygen } &= \ \text{3.0 moles}\cr
\text{1.0 moles of Al will need oxygen } &= \ \frac{3}{4}\cr
\text{0.3 moles of Al will need oxygen } &= \ \frac{3}{4} \ \times \ 0.3 \cr
&= \text{ 0.225 moles} \cr
\text{Molar mass of oxygen } \mathrm{(O_2)} &= \ \mathrm{32 \ gmol^{-1}} \cr
\text{Mass of 0.224 moles of oxygen } \mathrm{(O_2)} \text{ molecules} &= 32 \ \times \ 0.225 \cr
&= \ \mathrm{7.2g}
\end{aligned}}$$
Example 4
Aluminium metal reacts with oxygen to produce aluminium oxide. How many grams of oxygen will be required to react completely with 0.3 moles of aluminium?
$$
\displaylines{
\text{The balanced chemical equation for the reaction:}
\cr
\underset{\substack{\text{2 moles} \\ \text{4 g}}}{2\,\text{H}_{2\text{(g)}}}
+
\underset{\substack{\text{1 mole} \\ \text{32 g}}}{\text{O}_{2\text{(g)}}}
\rightarrow
\underset{\substack{\text{2 moles} \\ \text{36 g}}}{2\,\text{H}_2\text{O}_{\text{(l)}}}
\cr
\cr
\begin{aligned}
\text{4g of hydrogen produce } \mathrm{H_2O} \ &= \mathrm{36g \ of \ H_2O}\cr
\text{5g of hydrogen will produce } \mathrm{H_2O} &= \frac{36}{4} \ \times \ 5 \cr
&= \mathrm{45g\ of\ H_2O}\cr
\cr
\text{18g (1 mole) of } \mathrm{H_2O} \text{ contains molecules } &= \ 6.022 \times 10^{23}\cr
\text{36g of } \mathrm{H_2O} \text{ contains molecules } &= \ 6.022 \times 10^{23} \times \ 2\cr
&= 12.04\ \times\ 10^{23}\cr
\text{45g of } \mathrm{H_2O} \text{ contains molecules } &= \frac{45}{36} \ \times \ 12.04 \times 10^{23} \cr
&= 1.505 \ \times \ 10^{24} \ \text{molecules}
\end{aligned}
}
$$