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): \(n_{Na_3PO_4}=2.n_{P_2O_5}=2.0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{Na_3PO_4}=0,2.164=32,8\left(g\right)\)
\(\Rightarrow C_{\%_{Na_3PO_4}}=\dfrac{32,8}{764,2}.100\%=4,29\%\)
