\(\begin{array}{l}
P_2O_5+3H_2O\to 2H_3PO_4\\
KOH+H_3PO_4\to KH_2PO_4+H_2O(*)\\
2KOH+H_3PO_4\to K_2HPO_4+2H_2O(**)\\
3KOH+H_3PO_4\to K_3PO_4+3H_2O(***)\\
Dat\,(n_{KH_2PO_4};n_{K_2HPO_4};n_{K_3PO_4})=(x;y;z)(mol)\\
200ml=0,2l\\
n_{KOH}=0,2.1=0,2(mol)\\
n_{H_3PO_4}=\frac{39,2.5\%}{98}=0,02(mol)\\
n_{P_2O_5}=\frac{m}{142}(mol)\\
Theo\,PT:\,\sum n_{H_3PO_4}=\frac{m}{71}+0,02(mol)\\
TH_1:Xay\,ra\,(*),(**)\\
\to \begin{cases} x+2y=0,2\\ 136x+174y=12,6 \end{cases}\to Nghiem\,am\\
TH_2:Xay\,ra\,(**),(***)\\
\to \begin{cases} 2y+3z=0,2\\ 174y+212z=12,6 \end{cases}\to Nghiem\,am\\
TH_3:Xay\,ra\,(*)\to H_3PO_4\,du\,\frac{m}{71}-0,18(mol)\\
\to n_{KH_2PO_4}=0,2(mol)\\
\to m_{KH_2PO_4}=0,2.136=27,2>12,6(g)\to Loai\\
TH_4:Xay\,ra\,(**)\to KOH\,du\,0,14-\frac{3m}{71}(mol)\\
\to n_{K_3PO_4}=\frac{m}{71}+0,02(mol)\\
\to 56\left(0,14-\frac{3m}{71}\right)+212\left(\frac{m}{71}+0,02\right)=12,6\\
\to m\approx 0,839(g)
\end{array}\)

Đáp án:

 `m≈0,84(g)`

Giải thích các bước giải:

`n_{H_3PO_4}=\frac{39,2.5%}{98.100%}=0,02(mol)`

`n_{P_2O_5}=\frac{m}{142}(mol)`

`n_{KOH}=0,2.1=0,2(mol)`

$12,6(g)Y\begin{cases} K^{+}:0,2(mol)\\PO_4^{3-}:0,02+2.\frac{m}{142}(mol)\\OH^{-}:0,2-3.(0,02+2.\frac{m}{142})(mol) \end{cases}$

`⇔12,6=0,2.39+(0,02+2.\frac{m}{142}).95+[(0,2-3.(0,02+2.\frac{m}{142})].17`

`⇔m≈0,84(g)`