Some Basic Concepts of Chemistry - Result Question 10
####10. For the following reaction, the mass of water produced from $445 g$ of $C _{57} H _{110} O _6$ is :
$$ 2 C _{57} H _{110} O _6(s)+163 O _2(g) \rightarrow 114 CO _2(g)+110 H _2 O(l) $$
(2019 Main, 9 Jan II)
(a) $490 g$
(b) $495 g$
(c) $445 g$
(d) $890 g$
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Solution:
- $2 C _{57} H _{110} O _6(s)+163 O _2(g) \longrightarrow 110 H _2 O(l)+114 CO _2(g)$
Molecular mass of $C _{57} H _{110} O _6$ $=2 \times(12 \times 57+1 \times 110+16 \times 6) g=1780 g$
Molecular mass of $110 H _2 O=110(2+16)=1980 g$
$1780 g$ of $C _{57} H _{110} O _6$ produced $=1980 g$ of $H _2 O$.
$$ =495 of H _2 O $$