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PTHH:
\(4P+5O_2\overset{t^o}{--->}2P_2O_5\left(1\right)\)
\(P_2O_5+6NaOH--->2Na_3PO_4+3H_2O\left(2\right)\)
Ta có: \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)
Theo PT(1): \(n_{P_2O_5}=\dfrac{1}{2}.n_P=\dfrac{1}{2}.0,1=0,1\left(mol\right)\)
\(\Rightarrow m_{P_2O_5}=0,1.142=14,2\left(g\right)\)
Theo PT(2): \(n_{NaOH}=6.n_{P_2O_5}=6.0,1=0,6\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,6.40=24\left(g\right)\)
Ta có: \(C_{\%_{NaOH}}=\dfrac{24}{m_{dd_{NaOH}}}.100\%=3,2\%\)
\(\Rightarrow m_{dd_{NaOH}}=750\left(g\right)\)
\(\Rightarrow m_{dd_{Na_3PO_4}}=750+14,2=764,2\left(g\right)\)
Theo PT(2): \(n_{Na_3PO_4}=2.n_{P_2O_5}=2.0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{Na_3PO_4}=0,2.164=32,8\left(g\right)\)
\(\Rightarrow C_{\%_{Na_3PO_4}}=\dfrac{32,8}{764,2}.100\%=4,29\%\)
\(2PH_3+4O_2-^{t^o}\rightarrow P_2O_5+3H_2O\\ P_2O_5+3H_2O\rightarrow2H_3PO_4\\BTNT\left(P\right): n_{H_3PO_4}=n_{PH_3}=\dfrac{3,4}{34}=0,1\left(mol\right)\\ n_{KOH}=0,1.1,5=0,15\left(mol\right)\\ Tacó:\dfrac{n_{KOH}}{n_{H_3PO_4}}=\dfrac{0,15}{0,1}=1,5\Rightarrow Xảyracácpứ:\\ H_3PO_4+KOH\rightarrow KH_2PO_4+H_2O\\ H_3PO_4+2KOH\rightarrow K_2HPO_4+2H_2O\)
Gọi x,y lần lượt là số mol \(KH_2PO_4vàK_2HPO_4\)
=> \(\left\{{}\begin{matrix}x+y=0,1\left(BTNT\left(P\right)\right)\\x+2y=0,15\left(BTNT\left(K\right)\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,05\end{matrix}\right.\)
=> \(CM_{KH_2PO_4}=CM_{K_2HPO_4}=\dfrac{0,05}{0,1}=0,5M\)