Ta có : \(n_{Na2O}\frac{15,5}{62}0,25\left(mol\right)\)
\(PTHH:Na2O+H2O\rightarrow2NaOH\)(1)
\(2NaOH+H2SO4\rightarrow Na2SO4+2H2O\)(2)
Theo PT1: \(\Rightarrow n_{NaOH}=2.0,25=0,5\left(mol\right)\)
\(\Rightarrow CM_{NaOH}=\frac{0,5}{0,5}=1\)
Theo PT2:\(\Rightarrow n_{H2SO4}=0,25\left(mol\right)\Rightarrow m_{H2SO4}=24,5\left(g\right)\)
\(\Rightarrow m_{dd}=122,5\left(g\right)\Rightarrow V_{dd}=\frac{122,5}{1,14}=107,46\)
Na2O + H2O--->2NaOH(1)
2NaOH + H2SO4--->Na2SO4 +2H2O(2)
Ta có
n\(_{Na2O}=\)\(\frac{15,5}{62}=0,25\left(mol\right)\)
Theo pthh1
n\(_{NaOH}=2n_{Na2O}=0,5\left(mol\right)\)
C\(_{M\left(NaOH\right)}=\frac{0,5}{0,5}=1\left(M\right)\)
Theo pthh2
n\(_{H2SO4}=\frac{1}{2}n_{NaOH}=0,25\left(mol\right)\)
m\(_{H2SO4}=0,25.98=24,5\left(g\right)\)
m\(_{dd}=\frac{24,5.100}{20}=122,5\left(g\right)\)
V\(_{H2SO4}=\frac{122,5}{11.2}=\)109,375(l)
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