# Question #bb490

Jun 10, 2017

By the equation $C + \frac{1}{2} {O}_{2} \to C O$

If Whole 12g crbon is converted to $C O$ then total amount of CO produced would be 28g and the oxygen consumed would be 16g.So residual 4g should go to transform $C O \to C {O}_{2}$ on further oxidation of $C O$ by following reaction.

$C O + \frac{1}{2} {O}_{2} \to C {O}_{2}$

So residual 4g Oxygen will produce $\frac{44}{16} \times 4 = 11 g$ $C {O}_{2}$ and thereby consuming $\frac{28}{16} \times 4 = 7 g$ $C O$ leaving $\left(28 - 7\right) = 21 g$ $C O$.

So the ratio of $C O \mathmr{and} C {O}_{2}$ produced on full consumption of given 12g Carbon and 20g Oxgen would be $21 : 11$

NB

Calculation may be done in another way

First the whole 20g oxygen is consumed forming $C {O}_{2}$ first following equation $C + {O}_{2} \to C {O}_{2}$. For this 2og Oxygen will produce $\frac{44}{32} \times 20 g = 27.5 g$ $C {O}_{2}$ and this will consume
$\frac{12}{32} \times 20 g = 7.5 g$ $C$. The residual Carbon $\left(12 - 7.5\right) g = 4.5 g$ will now reduce $C {O}_{2} \to C O$ following the equation
$C {O}_{2} + C \to 2 C O$. So residual $4.5 g$ carbon will produce$\frac{2 \times 28}{12} \times 4.5 g = 21 g C O$ consuming $\frac{44}{12} \times 4.5 g = 16.5 g$ $C {O}_{2}$.
Thus the final amount of $C {O}_{2}$ in the mixture becomes

$\left(27.5 - 16.5\right) = 11 g$

So the ratio of $C O \mathmr{and} C {O}_{2}$ produced on full consumption of given 12g Carbon and 20g Oxgen would be $21 : 11$

Finally by ratio proportion

$\text{Compound"" "" "Amount" "" "Carbon" "" ""Oxygen}$

$C O \text{ "" "" "" "" "28g" "" "" "" "12g" " """ "" } 16 g$

$C {O}_{2} \text{ "" "" "" "" "44g" "" "" "" "12g" " """ "" } 16 g$

$C O \text{ "" "" "" "" "xg" "" "" "" "(12xx x)/28g" " """ } \frac{16 \times x}{28} g$

$C {O}_{2} \text{ "" "" "" "" "yg" "" "" "" "(12xx y)/44g" " """ } \frac{32 \times y}{44} g$

By the problem

$\frac{\frac{12 \times x}{28} + \frac{12 \times y}{44}}{\frac{16 \times x}{28} + \frac{32 \times y}{44}} = \frac{12}{20}$

$\implies \frac{x}{y} = \frac{21}{11}$