Đáp án:
a. Mg + 2HCl $→$ MgCl2 + H2
b. $n_{Mg}=\dfrac{2,8}{24}=\dfrac{7}{60}(mol)$
Theo PTHH $→n_{H_2}=n_{Mg}=\dfrac{7}{60}(mol)$
$→m_{H_2}=\dfrac{7}{60}\times 2=\dfrac{7}{30}(g)$
c. $n_{HCl}=2n_{H_2}=\dfrac{7}{30}(mol)$
$→m_{muoi}=2,8+\dfrac{7}{30}\times 35,5=11,0834(g)$