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
    $0.7$
  • $0.1$

Answer

Correct option: D.
$0.1$
d
$\mathop {3{\text{CaB}}{{\text{r}}_2}}\limits_{0.5}  + \mathop {2{{\text{K}}_3}{\text{P}}{{\text{O}}_4}}\limits_{0.2}  \to \mathop {{\text{C}}{{\text{a}}_3}{{\left( {{\text{P}}{{\text{O}}_4}} \right)}_2}}\limits_0  + 6{\text{KBr}}$

Acc. to $L.R.$                            $0.1$

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