a)\(Mg+2HCl\xrightarrow[]{}MgCl_2+H_2\)
b)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
\(m_{HCl}=\dfrac{100.3,65}{100\%}=3,65\left(g\right)\\ n_{HCl}=\dfrac{3,65}{365}=0,1\left(mol\right)\)
\(PTHH:Mg+2HCl\xrightarrow[]{}MgCl_2+H_2\)
Theo pt⇒\(\dfrac{0,2}{1}>\dfrac{0,1}{1}\Rightarrow Mg\) dư
\(PTHH:Mg+2HCl\xrightarrow[]{}MgCl_2+H_2\)
tỉ lệ : 1 2 1 1
số mol : 0,1 0,2 0,1 0,1
\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c)\(m_{ddMgCl_2}=\left(0,1.24\right)+100-\left(0,1.2\right)=102,2\left(g\right)\)
\(m_{MgCl_2}=0,1.95=9,5\left(g\right)\)
\(C\%_{MgCl_2}=\dfrac{9,5}{102,2}.100\%=9,3\%\)
