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
    $0.7$
  • D
    $0.1$

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