Mg+2HCl->MgCl2+H2
0,2--0,4---------0,2
nMg=4,8\24=0,2 mol
=>VH2=0,4.22,4=8,96l
=>mHCl=0,2.95=19g
=>mdd=130g
Mg+2HCl->MgCl2+H2
0,2--0,4---------0,2----0,2
nMg=4,8\24=0,2 mol
=>VH2=0,2.22,4=4,48l
=>mHCl=0,2.95=19g
=>mdd=130g
=>C%=0,2.95\130.100=14,61%
=>m đ muối =4,8+19-0,4=23,4g
=>C%=0,2.
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
a) Ta có: \(n_{Mg}=\frac{4,8}{24}=0,2\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,4mol\\n_{H_2}=0,2mol\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{ddHCl}=m=\frac{0,4\cdot36,5}{14,6\%}=100\left(g\right)\\V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\end{matrix}\right.\)
b) Theo PTHH: \(n_{Mg}=n_{MgCl_2}=0,2mol\) \(\Rightarrow m_{MgCl_2}=0,2\cdot95=19\left(g\right)\)
Ta có: \(m_{H_2}=0,2\cdot2=0,4\left(g\right)\)
Mặt khác: \(m_{dd}=m_{Mg}+m_{ddHCl}-m_{H_2}=104,6\left(g\right)\)
\(\Rightarrow C\%_{MgCl_2}=\frac{19}{104,6}\cdot100\approx18,16\%\)
