\(a,n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
0,3<----0,6<----0,3<-------0,3
\(\rightarrow\left\{{}\begin{matrix}m_{Mg}=0,3.24=7,2\left(g\right)\\m_{MgO}=15,2-7,2=8\left(g\right)\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{7,2}{15,2}.100\%=47,37\%\\\%m_{MgO}=100\%-47,37\%=52,63\%\end{matrix}\right.\)
\(b,n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)
PTHH: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)
0,2------>0,4------>0,2
\(n_{HCl}=0,4+0,6=1\left(mol\right)\\ \rightarrow V_{ddHCl}=\dfrac{1}{2}=0,5\left(l\right)\)
c, đề yêu cầu khối lượng chất rắn thu được khi đun nóng kết tủa đúng không?
\(n_{MgCl_2}=0,2+0,3=0,5\left(mol\right)\)
PTHH: \(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2\downarrow+2NaCl\)
0,5---------------------->0,5
\(Mg\left(OH\right)_2\xrightarrow[]{t^o}MgO+H_2O\)
0,5---------->0,5
\(\rightarrow m_{MgO}=0,5.40=20\left(g\right)\)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\
pthh:Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
0,3 0,6 0,3
\(m_{Mg}=0,3.24=7,2\left(g\right)\\ m_{MgO}=15,2-7,2=8\left(g\right)\\ \%m_{Mg}=\dfrac{7,2}{15,2}.100\%=47,36\%\\ \%m_{MgO}=10\%-47,36\%\\ n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\\
pthh:MgO+2HCl\rightarrow MgCl_2+H_2O\)
0,2 0,4
\(\Sigma n_{HCl}=0,4+0,6=1\left(mol\right)\)
\(V_{HCl}=\dfrac{1}{2}=0,5\left(l\right)\)
a. \(n_{H_2\left(đktc\right)}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
1 : 1 (mol)
0,3 : 0,3 (mol)
\(m_{Mg}=0,3.24=7,2\left(g\right)\)
\(\%m_{Mg}=\dfrac{7,2.100}{15,2}\%\approx47,37\%\)
\(\%m_{MgO}=\dfrac{\left(15,2-7,2\right).100}{15,2}\%\approx52,63\%\)
b. \(n_{MgO}=\dfrac{15,2-7,2}{40}=0,2\left(mol\right)\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
1 : 2 (mol)
0,2 : 0,4 (mol)
\(n_{HCl}=0,4+\dfrac{0,3.2}{1}=1\left(mol\right)\)
\(V_{ddHCl}=\dfrac{1}{2}=0,5\left(l\right)=500\left(ml\right)\)
