ta có: \(n_{P_2O_5}=\dfrac{76,5}{142}\approx0,54\left(mol\right)\)
PTHH: P2O5 + 3H2O ---> 2H3PO4.
Theo PT: \(n_{H_3PO_4}=2.n_{P_2O_5}=2.0,54=1,08\left(mol\right)\)
=> \(m_{H_3PO_4}=1,08.98=105,84\left(g\right)\)
=> C% = \(\dfrac{105,84}{500}.100\%=21,168\%\)