Bài 5:
a, PT: \(Na_2O+H_2O\rightarrow2NaOH\)
Ta có: \(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)
Theo PT: \(n_{NaOH}=2n_{Na_2O}=0,5\left(mol\right)\)
\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,5}{0,5}=1\left(M\right)\)
b, PT: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
Theo PT: \(n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,25.98=24,5\left(g\right)\Rightarrow m_{ddH_2SO_4}=\dfrac{24,5}{20\%}=122,5\left(g\right)\)
\(\Rightarrow V_{ddH_2SO_4}=\dfrac{122,5}{11,4}\approx10,75\left(ml\right)\)
c, Theo PT: \(n_{Na_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,25\left(mol\right)\)
\(\Rightarrow C_{M_{Na_2SO_4}}=\dfrac{0,25}{0,5+0,01075}\approx0,49\left(M\right)\)
Bài 6:
- Với dd HCl:
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
\(NaOH+HCl\rightarrow NaCl+H_2O\)
\(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\)
- Với dd H2SO4.
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
\(Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+CO_2+H_2O\)
