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Trả lời:
$Mg+2HCl\rightarrow MgCl_2+H_2$
$a, n_{Mg}=\dfrac{4,8}{24}=0,2\,(mol)$
Theo pt $⇒n_{H_2}=n_{Mg}=0,2\,(mol)$
$⇒V_{H_2}=0,2.22,4=4,48\,(l)$
$b,$ Theo pt
$⇒n_{HCl}=2.n_{Mg}=0,4\,(mol)$
$⇒m_{HCl}=0,4.36,5=14,6\,(g)$
$⇒C\%=\dfrac{14,6.100\%}{250}=5,84\%$.