\(n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=n_{H_2}=0,1mol\\ V_{H_2,đktc}=0,1.22,4=2,24l=2440ml\\ V_{H_2,đkc}=0,1.24,79=2,479l=24790ml\)

mol Fe= 5,6:56=0,1(mol)
Fe + 2HCl => FeCl2 + H2
cứ 1mol Fe sẽ ra 1 mol H2
suy ra mol H2=0,1(mol)
suy ra  V H2= mol.22,4= 0,1.22,4= 2,24 lít= 2240 (ml)
 

\(n_{C_2H_4Br_2}=\dfrac{28,2}{188}=0,15mol\\ a)C_2H_4+Br_2\xrightarrow[]{}C_2H_4Br_2\\ b)n_{C_2H_4}=n_{C_2H_4Br_2}=0,15mol\\ \%V_{C_2H_4}=\dfrac{0,15.22,4}{5,6}\cdot100\%=60\%\\ \%V_{CH_4}=100\%-60\%=40\%\)

chọn A

A nhé

\(a)n_{C_2H_4Br_2}=\dfrac{37,6}{188}=0,2mol\\ C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ n_{Br_2}=n_{C_2H_4}=n_{C_2H_4Br_2}=0,2mol\\ m_{Br_2}=0,2.160=32g\\ b)\%m_{C_2H_4}=\dfrac{0,2.28}{17,6}\cdot100\%=31,82\%\\ \%m_{CH_4}=100\%-31,82\%=68,18\%\)

Ta có \(n_{O_2}=\dfrac{V}{24,79}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)

\(\Rightarrow m_{O_2}=n.M=0,2.32=6,4\left(g\right)\)

=> Chọn A

a

\(SO_3+2NaOH\rightarrow Na_2SO_4+H_2O\\ n_{SO_3}=\dfrac{8}{40}=0,2\left(mol\right)\Rightarrow n_{Na_2SO_4}=n_{SO_3}=0,2\left(mol\right);n_{NaOH}=2.0,2=0,4\left(mol\right)\\ m_{NaOH}=0,4.40=16\left(g\right);m_{Na_2SO_4}=142.0,2=28,4\left(g\right)\)

thầy chưa hiểu lắm đề cần gì?