If $0.50 \ mol$ of $CaCl_2$ is mixed with $0.20 \ mol$ of $Na_3PO_4$,the maximum number of moles of $Ca_3(PO_4)_2$ which can be formed,is

  • A
    $0.7$
  • B
    $0.5$
  • C
    $0.2$
  • D
    $0.1$

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$A + 2B \to I$
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