\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

\(n_{Mg}=n_{H_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)

\(m_{Mg}=0.05\cdot24=1.2g\)

\(m_{MgO}=9.2-1.2=8\left(g\right)\)

a)

$Mg + 2HCl \to MgCl_2 + H_2$
$MgO + 2HCl \to MgCl_2 + H_2O$
b)

Theo PTHH :

$n_{Mg} = n_{H_2} = \dfrac{1,12}{22,4} = 0,05(mol)$

$m_{Mg} = 0,05.24 = 1,2(gam)$
$m_{MgO} = 9,2 - 1,2 = 8(gam)$

a) PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)

                   x_____2x_____________x                    (mol)

               \(Al_2\left(CO_3\right)_3+6HCl\rightarrow2AlCl_3+3H_2O+3CO_2\uparrow\)

                              y_____6y______________________3y             (mol)

Ta lập HPT: \(\left\{{}\begin{matrix}56x+234y=29\\x+3y=\dfrac{8,96}{22,4}=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=56\cdot0,1=5,6\left(g\right)\\m_{Al_2\left(CO_3\right)_3}=23,4\left(g\right)\end{matrix}\right.\)

b) Theo PTHH: \(n_{HCl}=2n_{Fe}+6n_{Al_2\left(CO_3\right)_3}=0,8\left(mol\right)\) \(\Rightarrow V_{ddHCl}=\dfrac{0,8}{2}=0,4\left(l\right)=400\left(ml\right)\)

c) Kết tủa sau phản ứng không có Al(OH)₃

Bảo toàn Sắt: \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=0,05\left(mol\right)\) \(\Rightarrow m_{Fe_2O_3}=0,05\cdot160=8\left(g\right)\)

nH2=0,1(mol)

PTHH: Mg + 2 HCl -> MgCl2 + H2

0,1__________0,2___________0,1(mol)

MgO + 2 HCl -> MgCl2 + H2O

0,05____0,1___0,05(mol)

mMg=0,1. 24= 2,4(g) -> mMgO=4,4-2,4= 2(g) -> nMgO=0,05((mol)

b) %mMg= (2,4/4,4).100=54,545%

=> %mMgO=45,455%

c) nHCl=0,3(mol) -> mHCl=0,3.36,5=10,95(g)

=> mddHCl=(10,95.100)/7,3=150(g)

\(\text{Đặt }\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\\ n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ a,PTHH:\left\{{}\begin{matrix}2Al+6HCl\rightarrow2AlCl_3+3H_2\\Fe+2HCl\rightarrow FeCl_2+H_2\end{matrix}\right.\\ b,\text{Theo đề ta có HPT: }\left\{{}\begin{matrix}27x+56y=8,3\\\dfrac{3}{2}x+y=0,25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\%_{Al}=\dfrac{0,1\cdot27}{8,3}\approx32,53\%\\\%_{Fe}\approx67,47\%\end{matrix}\right.\)

\(c,\left\{{}\begin{matrix}n_{AlCl_3}=0,1\left(mol\right)\\n_{FeCl_2}=0,1\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{AlCl_3}=0,1\cdot133,5=13,35\left(g\right)\\m_{FeCl_2}=0,1\cdot127=12,7\left(g\right)\end{matrix}\right.\\ \Rightarrow\sum m_{muối}=13,35+12,7=26,05\left(g\right)\)

Câu 4 : 

\(n_{H2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

a) Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)

             1         2              1           1

           0,05     0,1                       0,05

            \(MgO+2HCl\rightarrow MgCl_2+H_2O|\)

                1           2              1            1

               0,2        0,4

b) \(n_{Mg}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)

\(m_{Mg}=0,05.24=1.2\left(g\right)\)
\(m_{MgO}=9,2-1,2=8\left(g\right)\)

c) Có : \(m_{MgO}=8\left(g\right)\)

\(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)

\(n_{HCl\left(tổng\right)}=0,1+0,4=0,5\left(mol\right)\)

\(m_{HCl}=0,05.36,5=18,25\left(g\right)\)

\(m_{ddHCl}=\dfrac{18,25.100}{14,6}=125\left(g\right)\)

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