Làm câu a, các câu còn lại ttự, nếu rảnh mk làm nốt:
a) Số mol SO3(đktc):
.....................nSO3=13,44/22,4=0,6(mol)
PTHH: 2NaOH+ SO3----> Na2SO4+ H2O
.................1,2........0,6........................................(mol)
K/l dd NaOH 10%:
...................\(m_{dd}=\frac{m_{NaOH}.100}{10}=\frac{48.100}{10}=480\left(g\right)\)
\(a.n_{SO_3}=\frac{V}{22,4}=\frac{13,44}{22,4}=0,6\left(mol\right)\\ PTHH:SO_3+2NaOH\rightarrow Na_2SO_4+H_2O\\ \left(mol\right)1................2................1...........1\\ \left(mol\right)0,6..............1,2............0,6........0,6\\ m_{NaOH}=1,2.40=48\left(g\right)\\ \rightarrow m_{ddNaOH}=\frac{48.100}{10}=480\left(g\right)\)
\(b.m_{H_2SO_4}=\frac{200.9,8}{100}=19,6\left(g\right)\\ \rightarrow n_{H_2SO_4}=\frac{19,6}{98}=0,2\left(mol\right)\\ PTHH:H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\\ \left(mol\right)1................2................1...........2\\ \left(mol\right)0,2..............0,4............0,2........0,4\\ m_{NaOH}=0,2.40=8\left(g\right)\\ \rightarrow m_{ddNaOH}=\frac{8.100}{10}=80\left(g\right)\)
Sửa đề thành 16,1g ZnSO4 nhé!
\(c.n_{ZnSO_4}=\frac{16,1}{161}=0,1\left(mol\right)\\ PTHH:ZnSO_4+2NaOH\rightarrow Na_2SO_4+Zn\left(OH\right)_2\\ \left(mol\right)1................2................1...........1\\ \left(mol\right)0,1..............0,2............0,1........0,1\\ m_{NaOH}=0,1.40=4\left(g\right)\\ \rightarrow m_{ddNaOH}=\frac{4.100}{10}=40\left(g\right)\)
a) 2NaOH + SO3 → Na2SO4 + H2O (1)
\(n_{SO_3}=\frac{13,44}{22,4}=0,6\left(mol\right)\)
Theo pt1: \(n_{NaOH}=2n_{SO_3}=2\times0,6=1,2\left(mol\right)\)
\(\Rightarrow m_{NaOH}=1,2\times40=48\left(g\right)\)
\(\Rightarrow m_{ddNaOH}=\frac{48}{10\%}=480\left(g\right)\)
b) 2NaOH + H2SO4 → Na2SO4 + 2H2O (2)
\(m_{H_2SO_4}=200\times9,8\%=19,6\left(g\right)\)
\(\Rightarrow n_{H_2SO_4}=\frac{19,6}{98}=0,2\left(mol\right)\)
Theo pT2: \(n_{NaOH}=2n_{H_2SO_4}=2\times0,2=0,4\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,4\times40=16\left(g\right)\)
\(\Rightarrow m_{ddNaOH}=\frac{16}{10\%}=160\left(g\right)\)
