Bài 2:
a, \(n_P=\dfrac{7,75}{31}=0,25\left(mol\right)\)
PTHH: 4P + 5O2 ---to→ 2P2O5
Mol: 0,25 0,125
PTHH: P2O5 + 3H2O → 2H3PO4
Mol: 0,125 0,25
\(C\%_{ddH_3PO_4}=\dfrac{0,25.98.100\%}{280}=8,75\%\)
b, \(n_{NaOH}=0,3.2=0,6\left(mol\right)\)
Ta có: \(T=\dfrac{0,6}{0,25}=2,4\) ⇒ tạo ra 2 muối Na2HPO4, Na3PO4
PTHH: 2NaOH + H3PO4 → Na2HPO4 + 2H2O
Mol: 2x x x
PTHH: 3NaOH + H3PO4 → Na3PO4 + 3H2O
Mol: 3y y y
Ta có: \(\left\{{}\begin{matrix}x+y=0,25\\2x+3y=0,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,15\\y=0,1\end{matrix}\right.\)
\(m_{muối}=0,15.141+0,1.164=37,55\left(g\right)\)