$m_{ddH_{3}PO_{4}}=D.V=1,03.25=25,75$ (g)$m_{ctH_{3}PO_{4}}=\frac{6.25,75}{100}=1,545$ (g)
$n_{P_{2}O_{5}}=\frac{6}{142}$(mol)
Có: $2P_{2}O_{5} +6H_{2}O\rightarrow 4H_{3}PO_{4}$
$\Sigma m_{H_{3}PO_{4}}=\frac{6.4}{142.2}.98+ 1,545=9,827$ (g)
$C$%$_{H_{3}PO_{4}}$$=\frac{9,827}{6+25,75}.100=30,95$ %
Chúc bạn học tốt nha! ^^