Question

Cứuuu😞

Answer 1

Giúp mka đi mn mai ktra roii😞💔

Answer 2

\(n_{H_2}=\dfrac{11,2}{22,4}=0,5mol\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

0,5                                 0,5

\(m_{Fe}=0,5\cdot56=28\left(g\right)\Rightarrow\%m_{Fe}=\dfrac{28}{40}\cdot100\%=70\%\)

\(\Rightarrow\%m_{Cu}=100\%-70\%=30\%\)

Answer 3

\(Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ n_{Fe}=n_{H_2}=0,5\left(mol\right)\\ \%m_{Fe}=\dfrac{0,5.56}{40}.100=70\%\\ \%m_{Cu}=100-70=30\%\)