nMg = \(\dfrac{4,8}{24}\) = 0,2 mol
Mg +2 HCl -> MgCl2 + H2 \(\uparrow\)
0,2->0,4 ----->0,2------>0,2
=> VH2 = 0,2 . 22,4 = 4,48 (l)
=> CMHCl = \(\dfrac{0,4}{0,1}\) = 4M
=>mMgCl2 = 0,2 . 95 = 19 g
ta có: nMg= \(\dfrac{4,8}{24}\)= 0,2( mol)
PTPU
Mg+ 2HCl\(\rightarrow\) MgCl2+ H2\(\uparrow\)
0,2.....0,4.........0,2........0,2... (mol)
\(\Rightarrow\) VH2= 0,2. 22,4= 4,48( lít)
CM HCl= \(\dfrac{0,4}{0,1}\)= 4M
mMgCl2= 0,2. 95= 19( g)
Mg + 2HCl → MgCl2 + H2
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
Theo pT: \(n_{HCl}pư=2n_{Mg}=2\times0,2=0,4\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}pư=\dfrac{0,4}{0,1}=4\left(M\right)\)
Theo pT: \(n_{H_2}=n_{Mg}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,2\times22,4=4,48\left(l\right)\)
Theo PT: \(n_{MgCl_2}=n_{Mg}=0,2\left(mol\right)\)
\(\Rightarrow m_{MgCl_2}=0,2\times95=19\left(g\right)\)