Some Basic Concepts of Chemistry - Result Question 2
####2. The minimum amount of $O _2(g)$ consumed per gram of reactant is for the reaction (Given atomic mass : $Fe=56$, $O=16, Mg=24, P=31, C=12, H=1$ ) (2019 Main, 10 April II)
(a) $C _3 H _8(g)+5 O _2(g) \longrightarrow 3 CO _2(g)+4 H _2 O(l)$
(b) $P _4(s)+5 O _2(g) \longrightarrow P _4 O _{10}(s)$
(c) $4 Fe(s)+3 O _2(g) \longrightarrow 2 Fe _2 O _3(s)$
(d) $2 Mg(s)+O _2(g) \longrightarrow 2 MgO(s)$
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Solution:
$$ \begin{aligned} & \text { (a) } C _3 H _8(g)+\underset{44 g}{5 O _2(g)} \longrightarrow 3 CO _2(g)+4 H _2 O(l) \ & \Rightarrow 1 g \text { of reactant }=\frac{160}{44} g \text { of } O _2 \text { consumed }=3.64 g \end{aligned} $$
(b) $\underset{124 g}{P _4(s)}+\underset{160 g}{5 O _2(g)} \longrightarrow P _4 O _{10}(s)$
$$ \Rightarrow 1 g \text { of reactant }=\frac{160}{124} g \text { of } O _2 \text { consumed }=1.29 g $$
(c) $4 Fe(s)+3 O _2(g) \longrightarrow 2 Fe _2 O _3(s)$
$$ \Rightarrow 1 g \text { of reactant }=\frac{96}{224} g \text { of } O _2 \text { consumed }=0.43 g $$
(d) $\underset{48 g}{2 Mg(s)}+\underset{32 g}{O _2(g)} \longrightarrow 2 MgO(s)$
$$ \Rightarrow 1 g \text { of reactant }=\frac{32}{48} g \text { of } O _2 \text { consumed }=0.67 g $$
So, minimum amount of $O _2$ is consumed per gram of reactant $(Fe)$ in reaction $(c)$.
