\(n_{P_2O_5}=\dfrac{14.2}{142}=0.1\left(mol\right)\)
\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)
\(0.1..........................0.2\)
\(m_{H_3PO_4}=0.2\cdot98=19.6\left(g\right)\)
\(m_{dd}=14.2+180=194.2\left(g\right)\)
\(C\%H_3PO_4=\dfrac{19.6}{194.2}\cdot100\%=10.09\%\)