Some Basic Concepts of Chemistry 1 Question 2
2. The minimum amount of $\mathrm{O}_{2}(g)$ consumed per gram of reactant is for the reaction (Given atomic mass : $\mathrm{Fe}=56$, $\mathrm{O}=16, \mathrm{Mg}=24, \mathrm{P}=31, \mathrm{C}=12, \mathrm{H}=1$ ) (2019 Main, 10 April II)
(a) $\mathrm{C}{3} \mathrm{H}{8}(g)+5 \mathrm{O}{2}(g) \longrightarrow 3 \mathrm{CO}{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)$
(b) $\mathrm{P}{4}(s)+5 \mathrm{O}{2}(g) \longrightarrow \mathrm{P}{4} \mathrm{O}{10}(s)$
(c) $4 \mathrm{Fe}(s)+3 \mathrm{O}{2}(g) \longrightarrow 2 \mathrm{Fe}{2} \mathrm{O}_{3}(s)$
(d) $2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s)$
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Solution:
$$ \begin{aligned} & \text { (a) } \mathrm{C}{3} \mathrm{H}{8}(g)+\underset{44 \mathrm{~g}}{5 \mathrm{O}{2}(g)} \longrightarrow 3 \mathrm{CO}{2}(g)+4 \mathrm{H}{2} \mathrm{O}(l) \ & \Rightarrow 1 \mathrm{~g} \text { of reactant }=\frac{160}{44} \mathrm{~g} \text { of } \mathrm{O}{2} \text { consumed }=3.64 \mathrm{~g} \end{aligned} $$
(b) $\underset{124 \mathrm{~g}}{\mathrm{P}{4}(s)}+\underset{160 \mathrm{~g}}{5 \mathrm{O}{2}(g)} \longrightarrow \mathrm{P}{4} \mathrm{O}{10}(s)$
$$ \Rightarrow 1 \mathrm{~g} \text { of reactant }=\frac{160}{124} \mathrm{~g} \text { of } \mathrm{O}_{2} \text { consumed }=1.29 \mathrm{~g} $$
(c) $4 \mathrm{Fe}(s)+3 \mathrm{O}{2}(g) \longrightarrow 2 \mathrm{Fe}{2} \mathrm{O}_{3}(s)$
$$ \Rightarrow 1 \mathrm{~g} \text { of reactant }=\frac{96}{224} \mathrm{~g} \text { of } \mathrm{O}_{2} \text { consumed }=0.43 \mathrm{~g} $$
(d) $\underset{48 \mathrm{~g}}{2 \mathrm{Mg}(s)}+\underset{32 \mathrm{~g}}{\mathrm{O}_{2}(g)} \longrightarrow 2 \mathrm{MgO}(s)$
$$ \Rightarrow 1 \mathrm{~g} \text { of reactant }=\frac{32}{48} \mathrm{~g} \text { of } \mathrm{O}_{2} \text { consumed }=0.67 \mathrm{~g} $$
So, minimum amount of $\mathrm{O}_{2}$ is consumed per gram of reactant $(\mathrm{Fe})$ in reaction $(\mathrm{c})$.
