For the following reaction the mass of water produced from 445 g of $\mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_{6}$ is $2 \mathrm{C}_{57} \mathrm{H}_{110} \mathrm{O}_{6}(\mathrm{~s})+163 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 114 \mathrm{CO}_{2}(\mathrm{~g})+110 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})$

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Some Basic Concepts of Chemistry

For the following reaction the mass of water produced from 445 g of

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

2 C_{57} H_{110} O_{6} (s) + 163 O_{2} (g) ⟶ 114 CO_{2} (g) + 110 H_{2} O (l)

Answer: (C)

The balanced equation turns

2

moles of

C X_{57} H X_{110} O X_{6}

into

110

moles of water. The molar mass of

C X_{57} H X_{110} O X_{6}

57 (12) + 110 (1) + 6 (16) = 890 g/mol

, so

445 g

\frac{445}{890} = 0.5 mol

. By the

2 : 110

ratio this yields

\frac{110}{2} \times 0.5 = 27.5 mol

of water, and at

18 g/mol

the mass is

27.5 \times 18 = 495 g

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option