$\mathrm{X} \rightleftharpoons \mathrm{Y} ; \mathrm{K}_1=1.0$
$\mathrm{Y} \rightleftharpoons \mathrm{Z} ; \mathrm{K}_2=2.0$
$\mathrm{Z} \rightleftharpoons \mathrm{W} ; \mathrm{K}_3=4.0$
The equilibrium constant for the reaction $\mathrm{X} \rightleftharpoons \mathrm{W}$ is
$\mathrm{Y} \rightleftharpoons \mathrm{Z} \mathrm{k}_2=2$
$\mathrm{Z} \rightleftharpoons \omega \mathrm{k}_3=4$
$\mathrm{X} \rightleftharpoons \omega \mathrm{k}_1 \cdot \mathrm{k}_2 \cdot \mathrm{k}_3$
$\mathrm{k}=1 \times 2 \times 4$
$\mathrm{k}=8$
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$Ca_3(PO_4)_2 + SiO_2 + C + O_2 + H_2O \to CaSiO_3 + CO_2 + H_3PO_4$
If $1.0\ kg$ each of calcium phosphate and silica are used with excess of $C$ , $O_2$ and $H_2O$ , what is maximum quantity of phosphoric acid that can be produced. $(Ca = 40, P = 31, Si = 28)$
$A$ is
$1.\,\,(CH_3)_2 - \mathop C\limits^ + - CH_2 - CH_3$
$2.\,\,(CH_3)_3 - \mathop C\limits^ + $
$3.\,\,(CH_3)_2 - |\mathop C\limits^ + H|$
$4.\,\,CH_3 - \mathop C\limits^ + H_2$
$5.\,\,\mathop C\limits^ + H_3$