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Answer 1

Câu 3:

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

Ta có: \(n_{H_2}=\dfrac{0,336}{22,4}=0,015\left(mol\right)=n_{Fe}\) \(\Rightarrow m_{Fe}=0,015\cdot56=0,84\left(g\right)\)

Answer 2

Câu 4:

PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

Ta có: \(n_{H_2}=\dfrac{1,568}{22,4}=0,07\left(mol\right)=n_{Mg}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=0,07\cdot24=1,68\left(g\right)\\m_{Cu}=3-1,68=1,32\left(g\right)\end{matrix}\right.\)