$3Ba(OH)_2+P_2O_5\xrightarrow{}Ba_3(PO_4)_2+3H_2O$
$0,2$ : $\dfrac{0,2}{3}$ : $\dfrac{0,2}{3}$ : $0,2$ (mol)
500ml=0,5l
$n_{P_2O_5}=\dfrac{14,2}{142}=0,1(mol)$
$n_{Ba(OH)_2}=0,5.0,4=0,2(mol)$
Ta có: $\dfrac{n_{P_2O_5}}{1}>\dfrac{n_{Ba(OH)_2}}{3}(\dfrac{0,1}{1}>\dfrac{0,2}{3})$
=> $n_{P_2O_5}$ dư, dùng số mol của $Ba(OH)_2$
$m_{Ba_3(PO_4)_2}=\dfrac{0,2}{3}.601=≈40,07(g)$