Đặt \(n_{Mg}=x;n_{MgO}=y\)
\(PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\)
(mol) 1 2 1 1
(mol) x 2x x x
\(PTHH:MgO+2HCl\rightarrow MgCl_2+H_2O\)
(mol) 1 2 1 1
(mol) y 2y y y
\(hpt:\left\{{}\begin{matrix}24x+40y=4,4\\22,4x=2,24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}n_{Mg}=0,1\left(mol\right)\rightarrow m_{Mg}=0,1.24=2,4\left(g\right)\rightarrow\%m_{Mg}=\frac{2,4}{4,4}.100\%=54,54\left(\%\right)\\n_{MgO}=0,05\left(mol\right)\rightarrow m_{MgO}=0,05.40=2\left(g\right)\rightarrow\%m_{MgO}=100\%-54,54\%=45,46\left(\%\right)\end{matrix}\right.\)
\(m_{ddHCl}=\frac{36,5.\left(2.0,1+2.0,05\right).100\%}{14,6\%}=75\left(g\right)\)
\(m_{H_2}=x.2=0,1.2=0,2\left(g\right)\)
\(C\%_{ddspu}=\frac{\left(0,1+0,05\right).95}{4,4+75-0,2}.100\%=18\left(\%\right)\)
Mg + 2HCl → MgCl2 + H2 (1)
MgO + 2HCl → MgCl2 + H2O (2)
\(n_{H_2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
a) Theo Pt1: \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,1\times24=2,4\left(g\right)\)
\(\Rightarrow m_{MgO}=4,4-2,4=2\left(g\right)\)
\(\%m_{Mg}=\frac{2,4}{4,4}\times100\%=54,55\%\)
\(\%m_{MgO}=\frac{2}{4,4}\times100\%=45,45\%\)
b) \(m_{H_2}=0,1\times2=0,2\left(g\right)\)
Theo pT1: \(n_{HCl}=2n_{Mg}=2\times0,1=0,2\left(mol\right)\)
\(n_{MgO}=\frac{2}{40}=0,05\left(mol\right)\)
Theo Pt2: \(n_{HCl}=2n_{MgO}=2\times0,05=0,1\left(mol\right)\)
\(\Rightarrow\Sigma n_{HCl}=0,2+0,1=0,3\left(mol\right)\)
\(\Rightarrow\Sigma m_{HCl}=0,3\times36,5=10,95\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\frac{10,95}{14,6\%}=75\left(g\right)\)
Ta có: \(m_{dd}saupứ=4,4+75-0,2=79,2\left(g\right)\)
Theo pt1: \(n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\)
Theo Pt2: \(n_{MgCl_2}=n_{MgO}=0,05\left(mol\right)\)
\(\Rightarrow\Sigma n_{MgCl_2}=0,1+0,05=0,15\left(mol\right)\)
\(\Rightarrow\Sigma m_{MgCl_2}=0,15\times95=14,25\left(g\right)\)
\(\Rightarrow C\%_{MgCl_2}=\frac{14,25}{79,2}\times100\%=17,99\%\)
nH2 = 2.24/22.4 = 0.1 mol
Mg + 2HCl --> MgCl2 + H2
0.1___0.2______0.1____0.1
mMg = 2.4 g
=> mMgO = 4.4 - 2.4 = 2 g
=> nMgO = 0.05 mol
MgO + 2HCl --> MgCl2 + H2O
0.05____0.1______0.05
%Mg = 54.55%
%MgO = 45.45%
nHCl = 0.2 + 0.1 = 0.3 mol
mddHCL = 0.3*36.5*100/14.6=75 g
mdd sau phản ứng = 4.4 + 75 - 0.2 = 79.2 g
mMgCl2 = 0.15*95=14.25g
C%MgCl2 = 18%
