$\mathrm{X} \rightleftharpoons \mathrm{Y} ; \mathrm{K}_1=1.0$
$\mathrm{Y} \rightleftharpoons \mathrm{Z} ; \mathrm{K}_2=2.0$
$\mathrm{Z} \rightleftharpoons \mathrm{W} ; \mathrm{K}_3=4.0$
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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$Fe _{3} O _{4}( s )+4 CO ( g ) \rightarrow 3 Fe ( l )+4 CO _{2}( g )$
when $4.640 \,kg$ of $Fe _{3} O _{4}$ and $2.520 \,kg$ of $CO$ are allowed to react then the amount of iron (in $g$ ) produced is $....$
[Given : Molar Atomic mass $\left( g\, mol ^{-1}\right): Fe =56$
Molar Atomic mass $\left( g \,mol ^{-1}\right): 0=16$
Molar Atomic mass $\left( g\, mol ^{-1}\right):= C =12$
| Item $I$ | Item $II$ |
| $(a)$ Benzaldehyde | $(p)$ Mobile phase |
| $(b)$ Alumina | $(q)$ Adsorbent |
| $(c)$ Acetonitrile | $(r)$ Adsorbate |