Coi $n_{KNO_3} = n_{Ca(H_2PO_4)_2} = 1(mol)$
Ta có :
$m_X = 1.101 + 1.234 = 335(gam)$
$n_N = n_{KNO_3} = 1(mol)$
$n_{K_2O} = \dfrac{1}{2}n_{KNO_3} = 0,5(mol)$
$n_{P_2O_5} = n_{Ca(H_2PO_4)_2} = 1(mol)$
Suy ra :
$\%m_N = \dfrac{1.14}{335}.100\% = 4,18\%$
$\%m_{K_2O} = \dfrac{0,5.94}{335}.100\% = 14,03\%$
$\%m_{P_2O_5} = \dfrac{1.142}{335}.100\% = 42,39\%$