Question

Giứp mik câu 4 với, mik kẻm ơn ạ

Answer 1

Câu 4:

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{Fe}=n_{H_2}=0,2(mol)\\ \Rightarrow m_{Fe}=0,2.56=11,2(g)\\ \Rightarrow m_{Cu}=24-11,2=12,8(g)\\ \Rightarrow n_{Cu}=\dfrac{12,8}{64}=0,2(mol)\\ PTHH:Cu+2H_2SO_{4(đ)}\xrightarrow{t^o}CuSO_4+SO_2\uparrow+2H_2O\\ \Rightarrow n_{SO_2}=n_{Cu}=0,2(mol)\\ \Rightarrow V_{SO_2}=0,2.22,4=4,48(l)\)