Question

[Phan hoàng nhật vũ]

Giúp em câu 7 với 10

Answer 1

[Thảo Phương]

\(7.P_2O_5+3H_2O\rightarrow2H_3PO_4\\ n_{H_3PO_4}=2n_{P_2O_5}=0,2\left(mol\right)\\ Tacó:\dfrac{n_{KOH}}{n_{H_3PO_4}}=\dfrac{0,35}{0,2}=1,75\\ \Rightarrow Xảyracácphảnứng:\\ KOH+H_3PO_4\rightarrow KH_2PO_4+H_2O\\ 2KOH+H_3PO_4\rightarrow K_2HPO_4+2H_2O\\ Đặt:\left\{{}\begin{matrix}n_{KH_2PO_4}=x\left(mol\right)\\n_{KHPO_4}=y\left(mol\right)\end{matrix}\right.\\ Tacó:\left\{{}\begin{matrix}x+y=0,2\left(BTNT\left(P\right)\right)\\x+2y=0,35\left(BTNT\left(K\right)\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,15\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{KH_2PO_4}=6,8\left(g\right)\\m_{K_2HPO_4}=26,1\left(g\right)\end{matrix}\right.\)

Answer 2

[hưng phúc]

7. \(PTHH:P_2O_5+6KOH--->2K_3PO_4+3H_2O\)

Vậy chất sau phản ứng là K₃PO₄

Ta thấy: \(\dfrac{0,1}{1}>\dfrac{0,35}{6}\)

Vậy P₂O₅ dư.

Theo PT: \(n_{K_3PO_4}=\dfrac{1}{3}.n_{KOH}=\dfrac{1}{3}.0,35=\dfrac{7}{60}\left(mol\right)\)

\(\Rightarrow m_{K_3PO_4}=\dfrac{7}{60}.212\approx24,7\left(g\right)\)