Mg +H2SO4--->MgSO4 +2H2
a) Ta có
n\(_{H2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
Theo pthh
n\(_{Mg}=n_{H2}=0,1\left(mol\right)\)
m\(_{Mg}=0,1.24=2,4\left(g\right)\)
%m\(_{Mg}=\frac{2,4}{15,2}.100\%=15,79\%\)
%m\(_{Cu}=100-15,79=84,21\%\)
b) Theo pthh
n\(_{H2SO4}=n_{H2}=0,1\left(mol\right)\)
C\(_M=\frac{0,1}{0,5}=0,2\left(M\right)\)
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