Cho m(g) FeO,Fe2O3,Fe3O4 có số mol FeO bằng số mol Fe2O3 tác dụng đủ 200 (ml) dung dịch HCl 2(M). xác định m
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Đáp án C
Vì số mol của FeO bằng số mol của Fe2O3 nên ta coi hỗn hợp là Fe3O4
Ta có:
Fe3O4 +4H2SO4 → FeSO4 +Fe2(SO4)3 +4H2O
0,02 0,02
Trong 100 ml X sẽ có 0,01 mol FeSO4 nên:
10FeSO4 + 2KMnO4 + 8H2SO4 → 5Fe2(SO4)3 + K2SO4 + 2MnSO4 + 8H2O
→ nKMnO4 = 0,02 mol → VKMnO4 = 0,02/0,1 = 0,2 l
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Vì số mol FeO bằng Fe2O3.
\(FeO+Fe_2O_3\rightarrow Fe_3O_4\)
Quy đổi hỗn hợp chỉ gồm Fe3O4.
\(Fe_3O_4+8HCl\rightarrow FeCl_2+2FeCl_3+4H_2O\)
\(\Rightarrow n_{Fe3O4}=\frac{4,64}{56.3+16.4}=0,02\left(mol\right)\)
\(\Rightarrow n_{HCl}=8n_{Fe3O4}=0,16\left(mol\right)\)
\(\Rightarrow V_{HCl}=\frac{0,16}{1}=0,16\left(l\right)\)
Theo phản ứng: \(n_{H2O}=\frac{1}{2}n_{HCl}=0,08\left(mol\right)\)
BTKL:
\(m_{Fe3O4}+m_{HCl}=m_{muoi}+m_{H2O}\)
\(=4,64+0,16.36,5=m_{muoi}+0,08.18\)
\(\Rightarrow m_{muoi}=9,04\left(g\right)\)
Quy đổi hỗn hợp thành : Fe3O4
\(n_{HCl}=0.2\cdot2=0.4\left(mol\right)\)
\(Fe_3O_4+8HCl\rightarrow2FeCl_3+FeCl_2+4H_2O\)
\(0.05.........0.4\)
\(m=0.05\cdot232=11.6\left(g\right)\)
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