\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(n_{Mg}=0,1\left(mol\right)\)
\(m_{HCl}=\frac{150.10}{100}=15\left(g\right)\Rightarrow n_{HCl}=0,4\left(mol\right)\)
\(n_{Mg}< \frac{n_{HCl}}{2}\)
\(n_{H2}=n_{Mg}=0,1\left(mol\right)\Rightarrow V_{H2}=0,1.22,4=2,24\left(l\right)\)
\(m_{HCl\left(dư\right)}=15-\left(0,2.36,5\right)=7,7\left(g\right)\)
\(m_{MgCl2}=0,1.95=9,5\left(g\right)\)
\(m_{dd}=2,5+150-0,1.2=152,3\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{HCl\left(dư\right)}=\frac{7,7}{152,3}.100\%=5,06\%\\C\%_{MgCl2}=\frac{9,5}{152,5}.100\%=6,24\%\end{matrix}\right.\)
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