a) Mg +2HCl---->MgCl2 +H2
b) Ta có
n\(_{Mg}=\frac{4,8}{24}=0,2\left(mol\right)\)
Theo pthh
n\(_{H2}=n_{Mg}=0,2\left(mol\right)\)
V\(_{H2}=0,2.22,4=4,48\left(l\right)\)
c) Theo pthh
n\(_{MgCl2}=nMg=0,2\left(mol\right)\)
m\(_{MgCl2}=0,2.95=19\left(g\right)\)
d) Theo pthh
n\(_{HCl}=2n_{Mg}=0,4\left(mol\right)\)
C\(_{M\left(HCl\right)}=\frac{0,4}{0,02}=20M\)
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