ta có: nMg= \(\dfrac{1,2}{24}\)= 0,05( mol)
nHCl= 0,05.3= 0,15( mol)
PTPU
Mg+ 2HCl\(\rightarrow\) MgCl2+ H2
ta có tỉ lệ: \(\dfrac{0,05}{1}\)< \(\dfrac{0,15}{2}\)\(\Rightarrow\)HCl dư, Mg hết
theo PTPU ta có: nH2= nMg= 0,05( mol)
\(\Rightarrow\) VH2= 0,05.22,4= 1,12( lít)
nMgCl2= nMg= 0,05( mol)
nHCl pư= 2.nMg= 0,1( mol)
\(\Rightarrow\) nHCl dư= 0,15- 0,1= 0,05( mol)
\(\Rightarrow\) CM HCl dư= \(\dfrac{0,05}{0,05}\)= 1M
CM MgCl2= \(\dfrac{0,05}{0,05}\)= 1M
a) Mg + 2HCl → MgCl2 + H2
\(n_{Mg}=\dfrac{1,2}{24}=0,05\left(mol\right)\)
\(n_{HCl}=0,05\times3=0,15\left(mol\right)\)
Theo PT: \(n_{Mg}=\dfrac{1}{2}n_{HCl}\)
Theo bài: \(n_{Mg}=\dfrac{1}{3}n_{HCl}\)
Vì: \(\dfrac{1}{3}< \dfrac{1}{2}\) ⇒ HCl dư
b) Theo PT: \(n_{H_2}=n_{Mg}=0,05\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,05\times22,4=1,12\left(l\right)\)
c) Dung dịch sau phản ứng gồm: HCl dư và MgCl2
Theo PT: \(n_{HCl}pư=2n_{H_2}=2\times0,05=0,1\left(mol\right)\)
\(\Rightarrow n_{HCl}dư=0,15-0,1=0,05\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}dư}=\dfrac{0,05}{0,05}=1\left(M\right)\)
Theo PT: \(n_{MgCl_2}=n_{H_2}=0,05\left(mol\right)\)
\(\Rightarrow C_{M_{MgCl_2}}=\dfrac{0,05}{0,05}=1\left(M\right)\)
