Bài 1:
\(n_{H_2SO_4}=0,4.0,2=0,8\left(mol\right)\Rightarrow m_{H_2SO_4}=0,8.36,5=29,2\left(g\right)\)
PTHH: FeO + H2SO4 → FeSO4 + H2O
PTHH: Fe2O3 + 3H2SO4 → Fe2(SO4)3 + 3H2O
PTHH: Fe3O4 + 4H2SO4 → FeSO4 + Fe2(SO4)3 + 4H2O
Ta có: \(n_{H_2O}=n_{H_2SO_4}=0,8\left(mol\right)\Rightarrow m_{H_2O}=0,8.18=14,4\left(g\right)\)
Theo ĐLBTKL ta có:
\(m_{oxit}+m_{H_2SO_4}=m_{muối}+m_{H_2O}\)
\(\Leftrightarrow m_{muối}=m_{oxit}+m_{H_2SO_4}-m_{H_2O}=44,8+29,2-14,4=59,6\left(g\right)\)
Bài 2:
\(n_{HCl}=0,4.1=0,4\left(mol\right)\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\)
PTHH: Mg(OH)2 + 2HCl → MgCl2 + 2H2O
PTHH: Cu(OH)2 + 2HCl → CuCl2 + 2H2O
PTHH: NaOH + HCl → NaCl + H2O
Ta có: \(n_{HCl}=n_{H_2O}=0,4\left(mol\right)\Rightarrow m_{H_2O}=0,4.18=7,2\left(g\right)\)
Theo ĐLBTKL ta có:
\(m_{bazơ}+m_{HCl}=m_{muối}+m_{H_2O}\)
\(\Rightarrow m_{bazơ}=m_{muối}+m_{H_2O}-m_{HCl}=24,1+7,2-14,6=16,7\left(g\right)\)
a,\(n_{Al_2O_3}=\dfrac{5,1}{102}=0,05\left(mol\right)\)
PTHH: Al2O3 + 6HCl → 2AlCl3 + 3H2O
Mol: 0,05 0,3
\(V_{ddHCl}=\dfrac{0,3}{10}=0,03\left(l\right)\)
b,
PTHH: Al2O3 + 2NaOH → 2NaAlO2 + H2O
Mol: 0,05 0,1
\(V_{ddNaOH}=\dfrac{0,1}{10}=0,01\left(l\right)\)
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