\(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)
PTHH: 2Fe + 3Cl2 --to--> 2FeCl3
0,4-->0,6
MnO2 + 4HCl --> MnCl2 + Cl2 + 2H2O
2,4<-------------0,6
=> \(n_{NaCl}=2,4\left(mol\right)\)
=> \(\%NaCl=\dfrac{2,4.58,5}{200}.100\%=70,2\%\)
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