nH2 = 1,12 : 22,4 = 0,05 mol
Mg + 2HCl -> MgCl2 + H2 (1)
0,05 0,1 0,05 0,05 (mol)
=> mMg = 0,05 .24 = 1,2 g
%mMg = 1,2.100%\9,2=13,04%
=> mMgO = 9,2 - 1,2 = 8 g => nMgO = 0,2 mol
%mMgO = 100% - 13,04% = 86,96%
b) MgO + 2HCl -> MgCl2 + H2O (2)
0,2 0,4
Từ pt (1,2) => nHCl = 0,4 + 0,1 = 0,5 mol => mHCl = 18,25 g
mddHCl = 18,25.100\14,6=125g
\(PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\left(1\right)\)
_________0,05 ___________________0,05
\(MgO+2HCl\rightarrow MgCl_2+H_2O\left(2\right)\)
\(n_{khi}=\frac{1,12}{22,4}=0,05\left(mol\right)\)
Theo PTHH(1)
\(n_{Mg}=n_{H2}=0,05\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,05.24=1,2\left(g\right)\)
\(\Rightarrow m_{MgO}=9,2-1,2=8\left(g\right)\)
\(\Rightarrow\%m_{Mg}=\frac{1,2.100}{9,2}=13,04\%\)
\(\Rightarrow\%m_{MgO}=100\%-13,04\%=86,96\%\)
b)
\(n_{MgO}=\frac{8}{40}=0,2\left(mol\right)\)
Theo PTHH
\(n_{HCl\left(1\right)}=0,4\left(mol\right)\)
\(n_{HCl\left(2\right)}=0,1\left(mol\right)\)
\(\Rightarrow\Sigma n_{HCl}=0,5\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,5.36,5=18,25\left(g\right)\)
\(\Rightarrow m_{dd\left(HCl\right)}=\frac{18,25.100}{14,6}=125\left(g\right)\)
\(\Rightarrow m_{dd\left(spu\right)}=125+9,2-0,05.2=134,1\left(g\right)\)
Theo PTHH(1) và (2)
\(\Rightarrow n_{MgCk2}=0,25\left(mol\right)\)
\(\Rightarrow m_{MgCl2}=0,25.95=23,75\left(g\right)\)
\(\Rightarrow CM_{MgCl2}=\frac{23,75}{134,1}.100\%=17,71\%\)
