\(n_{H_2}=\dfrac{22,4}{22,4}=1mol\)
Mg+H2SO4\(\rightarrow\)MgSO4+H2
\(n_{Mg}=n_{H_2}=1mol\rightarrow m_{Mg}=24g\)
-Chất rắn B là Cu:
Cu+2H2SO4(đặc)\(\overset{t^0}{\rightarrow}\)CuSO4+SO2+2H2O
-Khí C là SO2
\(n_{Cu}=n_{SO_2}=\dfrac{3,36}{22,4}=0,15mol\)
mCu=0,15.64=9,6gam
%Mg=\(\dfrac{24}{24+9,6}.100\%\approx71,43\%\)
%Cu=100%-71,43%=28,57%