\(n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{H_2}=n_{Fe}=0,1mol\\ V_{H_2,đktc}=0,1.22,4=2,24l\\ V_{H_2,đkc}=0,1.24,79=2,479l\)
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