Question

If $0.50$ mole of monovalent metal ($M^{+1}$) halide is mixed with $0.2$ mole of a divalent metal ($L^{-2}$) phosphate, the maximum number of moles of $M_2{PO}_4$ that can be formed is:

• A. $0.25$
• B. $0.30$
• C. $0.16$
• D. $0.20$

Answer 1

The correct option is A $0.25$

$0.5$ mole of $MX$ is mixed with $0.2$ mole of $L_3{\left({PO}_4\right)}_2$ (divalent metal phosphate)

$L_3{\left({PO}_4\right)}_2\rightleftharpoons {2PO}_4^{3-}+3L^{2+}$

$\therefore$ $0.4$ mole of ${PO}_4^{3-}$ is present

For $1$ mole of $M_2{PO}_4$ formation needs $2$ moles of $M^{+1}$.

$\therefore \frac{0.5}{2}=0.25$ mole of $M_2{PO}_4$ will be formed.