a,\(n_{HCl}=0,2.1=0,2\left(mol\right)\)
PTHH: Fe + 2HCl → FeCl2 + H2
Mol: 0,1 0,2 0,1
\(\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)
b,\(C_{M_{ddFeCl_2}}=\dfrac{0,1}{0,2}=0,5M\)
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\(n_{H_2SO_4}=0,2.1=0,2\left(mol\right)\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ 0,2.......0,2.........0,2.......0,2\left(mol\right)\\ m=m_{Fe}=0,2.56=11,2\left(g\right)\\ b.V_{ddFeSO_4}=V_{ddH_2SO_4}=200\left(ml\right)=0,2\left(l\right)\\ C_{MddFeSO_4}=\dfrac{0,2}{0,2}=1\left(M\right)\)
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