a, \(Na_2O+2HCl\rightarrow2NaCl+H_2O\)
b, Ta có: \(n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\)
Theo PT: \(n_{NaCl}=2n_{Na_2O}=0,4\left(mol\right)\)
\(\Rightarrow C\%_{NaCl}=\dfrac{0,4.58,5}{12,4+200}.100\%\approx11,02\%\)
\(n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\)
a.
\(Na_2O+2HCl\rightarrow2NaCl+H_2O\)
0,2---------------->0,4
b.
\(C\%_{NaCl}=\dfrac{0,4.58,5.100\%}{12,4+200,}=11,02\%\)
`HaNa♬D`
\(a/Na_2O+2HCl\rightarrow2NaCl+H_2O\\ b/n_{Na_2O}=\dfrac{12,4}{62}=0,2mol\\ n_{NaCl}=2.0,2=0,4mol\\ C_{\%NaCl}=\dfrac{0,4.58,5}{200+12,4}\cdot100=11,02\%\)
