\(PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\left(1\right)\)
______0,05______0,1_______0,05____0,05
\(MgO+2HCl\rightarrow MgCl_2+H_2O\left(2\right)\)
__ 0,2___0,4___0,2 ________0,2
\(n_{H2}=\frac{1,12}{22,4}=0,05\left(mol\right)\)
Theo PTHH (1)
\(n_{Mg}=0,05\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,05.24=1,2\left(g\right)\)
\(\Rightarrow m_{MgO}=9,2-1,2=8\left(g\right)\)
\(n_{MgO}=\frac{8}{40}=0,2\left(mol\right)\)
\(\Sigma n_{HCl}=n_{HCl\left(1\right)}+n_{HCl\left(2\right)}=0,5\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,5.36,5=18,25\left(g\right)\)
\(m_{HCl\left(bđ\right)}=\frac{300.29,2}{100}=87,6\left(g\right)\)
\(\Rightarrow m_{HCl\left(dư\right)}=87,6-18,25=69,35\left(g\right)\)
\(\Sigma m_{MgCl2}=m_{MgCl2\left(1\right)}+m_{MgCl2\left(2\right)}=0,05.95+0,2.95=23,75\left(g\right)\)
\(m_{dd\left(spu\right)}=9,2+300=309,2\left(g\right)\)
\(C\%_{MgCl2}=\frac{23,75.100\%}{309,2}=7,68\%\)
\(C\%_{HCl}=\frac{69,35.100\%}{309,2}=22,43\%\)
