Q12 / 90 · JEE Main 2019 (09 Jan Shift 2) · Chemistry

For the following reaction, the mass of water produced from $445 g$ of $C_{57} H_{110} O_{6}$ is: $2 C_{57} H_{110} O_{6} (s) + 163 O_{2} (g) \to 114 {CO}_{2} (g) + 110 H_{2} O (l)$

A
$490 g$
B
$890 g$
C
$445 g$
Correct answer
D
$495 g$

Solution

C_{57} H_{110} O_{6} + \frac{163}{2} O_{2} \to 57 C O_{2} + 55 H_{2} O

Molecular weight of

C_{57} H_{110} O_{6} = 890

Moles of

C_{57} H_{110} O_{6} = \frac{w e i g h t (g m)}{M o l e c u l a r w e i g h t}

= \frac{445}{890} = \frac{1}{2}

According to stoichiometry

∵ 1 m o l e

of

C_{57} H_{110} O_{6}

gives

= 55 m o l e

of

H_{2} O

∵ \frac{1}{2} m o l e

of

C_{57} H_{110} O_{6}

gives

= 55 \times \frac{1}{2} m o l e

of

H_{2} O

Wt. of

H_{2} O =

mole of

H_{2} O \times M

Wt. of

H_{2} O

= \frac{1}{2} \times 55 \times 18 = 495 g m

Hence, the correct answer is option: 'D'.

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