Mg + 2HCl → MgCl2 + H2
\(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)
\(m_{HCl}=200\times7,3\%=14,6\left(g\right)\)
\(\Rightarrow n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
Theo PT: \(n_{Mg}=\dfrac{1}{2}n_{HCl}\)
Theo bài: \(n_{Mg}=\dfrac{3}{8}n_{HCl}\)
Vì \(\dfrac{3}{8}< \dfrac{1}{2}\) ⇒ HCl dư
Dung dịch sau phản ứng gồm: HCl dư, MgCl2
Theo PT: \(n_{HCl}pư=2n_{Mg}=2\times0,15=0,3\left(mol\right)\)
\(\Rightarrow n_{HCl}dư=0,4-0,3=0,1\left(mol\right)\)
\(\Rightarrow m_{HCl}dư=0,1\times36,5=3,65\left(g\right)\)
Theo PT: \(n_{MgCl_2}=n_{Mg}=0,15\left(mol\right)\)
\(\Rightarrow m_{MgCl_2}=0,15\times95=14,25\left(g\right)\)
\(\Sigma m_{ctan}saupư=m_{HCl}dư+m_{MgCl_2}=3,65+14,25=17,9\left(g\right)\)
\(\Sigma m_{dd}saupư=m_{Mg}+m_{ddHCl}=3,6+200=203,6\left(g\right)\)
\(\Rightarrow C\%_{dd}saupư=\dfrac{17,9}{203,6}\times100\%=8,792\%\)
