$n_{P_2O_5}=\dfrac{22,72}{142}=0,16(mol)$
$\Rightarrow n_{H_3PO_4}=2n_{P_2O_5}=0,32(mol)$
$n_{KOH}=\dfrac{100.33,6\%}{56}=0,6(mol)$
$\dfrac{n_{KOH}}{n_{H_3PO_4}}=1,875$
$\Rightarrow$ tạo $KH_2PO_4$ (x mol), $K_2HPO_4$ (y mol)
Bảo toàn K: $2x+y=0,6$
Bảo toàn P: $x+y=0,32$
$\Rightarrow x=0,28; y=0,04$
$m_{dd}=22,72+100=122,72g$
$C\%_{KH_2PO_4}=\dfrac{0,28.136.100}{122,72}=31,03\%$
$C\%_{K_2HPO_4}=\dfrac{0,04.174.100}{122,72}=5,67\%$