# Question c90a3

Jan 23, 2017

4C_2H_3Cl_3+11O_2→ 8CO_2+ 6H_2O + 6Cl_2#

Above equation suggests that the $C {O}_{2} \mathmr{and} C {l}_{2}$ will be produced in the mole ratio $8 : 6 = 4 : 3$

So ratio of their masses will be
$= \left(4 \times \text{molar mass of " CO_2)/(3xx"molar mass of } C {l}_{2}\right)$

$= \left(4 \cdot 44 g \text{/mol")/(3*71g"/mol}\right)$

$= \frac{176}{213}$

If xg $C {l}_{2}$ is produced on production of 59.5g $C {O}_{2}$ then we can write
$\frac{59.5}{x} = \frac{176}{213}$

$\implies x = \frac{59.5 \times 213}{176} g \approx 72 g$