Some Basic Concepts of Chemistry 1 Question 10
10. For the following reaction, the mass of water produced from $445 \mathrm{~g}$ of $\mathrm{C}{57} \mathrm{H}{110} \mathrm{O}_{6}$ is :
$$ 2 \mathrm{C}{57} \mathrm{H}{110} \mathrm{O}{6}(s)+163 \mathrm{O}{2}(g) \rightarrow 114 \mathrm{CO}{2}(g)+110 \mathrm{H}{2} \mathrm{O}(l) $$
(2019 Main, 9 Jan II)
(a) $490 \mathrm{~g}$
(b) $495 \mathrm{~g}$
(c) $445 \mathrm{~g}$
(d) $890 \mathrm{~g}$
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Solution:
- $2 \mathrm{C}{57} \mathrm{H}{110} \mathrm{O}{6}(s)+163 \mathrm{O}{2}(g) \longrightarrow 110 \mathrm{H}{2} \mathrm{O}(l)+114 \mathrm{CO}{2}(g)$
Molecular mass of $\mathrm{C}{57} \mathrm{H}{110} \mathrm{O}_{6}$ $=2 \times(12 \times 57+1 \times 110+16 \times 6) \mathrm{g}=1780 \mathrm{~g}$
Molecular mass of $110 \mathrm{H}_{2} \mathrm{O}=110(2+16)=1980 \mathrm{~g}$
$1780 \mathrm{~g}$ of $\mathrm{C}{57} \mathrm{H}{110} \mathrm{O}{6}$ produced $=1980 \mathrm{~g}$ of $\mathrm{H}{2} \mathrm{O}$.
$$ =495 \mathrm{of} \mathrm{H}_{2} \mathrm{O} $$
