$\left.\begin{array}{cc}0.8 g & x \text { cal } \\ 12 g & (?)\end{array}\right\}=\frac{12 x }{0.8}=15 x$

$ii)$ $C +\frac{1}{2} O _2 \rightarrow CO$

$\left.\begin{array}{cc}0.8 g & \text { y cal } \\ \therefore 12 g & \text { (?) }\end{array}\right\}=\frac{12 y }{0.8}=15 y$

$(i) - (ii):$ $CO +\frac{1}{2} O _2 \longrightarrow CO _2$

$\left.\begin{array}{r}28 \,g \rightarrow 15 x -15 y \\ \therefore 1.86\, g \rightarrow(?)\end{array}\right\}=\frac{1.86 \times 15( x - y )}{28}$

$=\frac{27.9(x-y)}{28}=x-y$