Answer
$V(H_3PO_4)=2ml$
Work Step by Step
Considering stoichiometric coefficients, we can conclude that $n(H_3PO_4):n(Mg(OH)_2)=2:3\implies n(H_3PO_4)=\frac{2}{3}n(Mg(OH)_2)\implies c(H_3PO_4)\times V(H_3PO_4) = \frac{2}{3}c(Mg(OH)_2)\times V(Mg(OH)_2)\implies V(H_3PO_4)=\frac{2\times c(Mg(OH)_2)\times V(Mg(OH)_2)}{3\times c(H_3PO_4)}=\frac{2\times 0.15\frac{mol}{dm^3}\times 0.045dm^3}{3\times 2.25\frac{mol}{dm^3}}=0.002dm^3=2ml$
