a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(n_{Mg}=\dfrac{m}{M}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
Theo PTHH : \(n_{H_2}=n_{Mg}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2}=n.22,4=0,2.22,4=4,48\left(l\right)\)
b, Theo PTHH : \(n_{HCl}=2.n_{Mg}=2.0,2=0,4\left(mol\right)\)
\(\Rightarrow m_{HCl}=n.M=0,4.36,5=14,6\left(g\right)\)
\(m_{ddHCl}=\dfrac{m_{ct}.100\%}{C_{\%}}=\dfrac{14,6.100\%}{7,3\%}=200\left(g\right)\)
c,\(m_{H_2}=n.M=0,2.2=0,4\left(g\right)\)
Theo ĐLBTKL ,có :
\(m_{MgCl_2}=m_{Mg}+m_{HCl}-m_{H_2}\)
hay \(m_{ddMgCl_2}=4,8+200-0,4=204,4\left(g\right)\)
Theo PTHH: \(n_{MgCl_2}=n_{Mg}=0,2\left(mol\right)\)
\(\Rightarrow m_{MgCl_2}=n.M=0,2.95=19\left(g\right)\)
\(C_{\%MgCl_2}=\dfrac{19}{204,4}.100\%\sim9,3\%\)
