MCQ
If $0.5 \,mol$ of $CaBr_2$ is mixed with $0.2\, mol$ of $K_3PO_4$ then the maximum number of moles of $Ca_3(PO_4)_2$ obtained will be :-
- A$0.5$
- B$0.2$
- C$0.7$
- ✓$0.1$
Acc. to $L.R.$ $0.1$
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$S\left( s \right) + {O_2}\left( g \right) \rightleftharpoons S{O_2}\left( g \right);{K_1} = {10^{52}}$
$2S\left( s \right) + 3{O_2}\left( g \right) \rightleftharpoons 2S{O_3}\left( g \right);{K_2} = {10^{129}}$
The equilibrium constant for the reaction $2S{O_2}\left( g \right) + {O_2}\left( g \right) \rightleftharpoons 2S{O_3}\left( g \right)$ is
