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 = \(\frac{1,2.100\%}{9,2}=13,04\%\)

=> 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 = \(\frac{18,25.100}{14,6}=125g\)

Ta có: \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

PTHH:

Mg + 2HCl ---> MgCl₂ + H₂ (1)

MgO + 2HCl ---> MgCl₂ + H₂O (2)

Theo PT₍₁₎: \(n_{Mg}=n_{H_2}=0,05\left(mol\right)\)

=> \(m_{Mg}=0,05.24=1,2\left(g\right)\)

=> \(\%_{m_{Mg}}=\dfrac{1,2}{9,2}.100\%=13,04\%\)

\(\%_{m_{MgO}}=100\%-13,04\%=86,96\%\)

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

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

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

\(0.05..............................0.05\)

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

\(m_{MgO}=9.5-1.2=8.3\left(g\right)\)

\(\%Mg=\dfrac{1.2}{9.5}\cdot100\%=12.63\%\)

\(\%MgO=100-12.63=87.36\%\)

a) mHCl= 14,6%. 125=18,25(g)

=> nHCl= 0,5(mol)

PTHH: Mg + 2 HCl -> MgCl2+ H2

x_______2x________x_____x(mol)

MgO + 2 HCl -> MgCl2 + H2O

y______2y_______y___y(mol)

mMg+ mMgO= 7,6

<=> 24x+40y=7,6(g) (a)

Mặt khác: nHCl(tổng)=0,5

<=>2x+2y=0,5 (b)

Từ (a), (b) , ta có hpt:

\(\left\{{}\begin{matrix}24x+40y=7,6\\2x+2y=0,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,15\\y=0,1\end{matrix}\right.\)

=> mMg=0,15.24=3,6(g)

=> \(\%mMg=\frac{3,6}{7,6}.100\approx47,368\%\\ \rightarrow\%mMgO\approx100\%-47,368\%\approx52,632\%\)

b) nMgCl2(tổng)=x+y=0,15+0,1=0,25(mol)

=> mMgCl2(tổng)=95.0,25=23,75(g)

mddMgCl2= m(MgO,Mg)+ mddHCl - mH2= 7,6+125-0,15.1=132,45(g)

=> \(C\%ddMgCl2=\frac{23,75}{132,45}.100\approx17,931\%\)

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 : \(m_{MgO}=8\left(g\right)\)

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

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

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

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

 Chúc bạn học tốt

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

0,05\(\leftarrow\) 0,05 (mol)

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

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

Theo ptpu: \(n_{Mg}=0,05\left(mol\right)\)

\(\Rightarrow m_{Mg}=0,05.24=1,2\left(g\right)\)

\(\Rightarrow\%Mg=\dfrac{1,2.100\%}{9,2}=13,04\%\)

\(\Rightarrow\%MgO=100\%-13,04\%=86,96\%\)

nH2 = 1,12/22,4 = 0,05 (mol)

Mg + 2HCl -> MgCl2 + H2

0,05..........0,1..................0,05 (mol)

MgO + 2HCl -> MgCl2 + H2O

=> m Mg = 0,05 . 24 = 1,2 (g)

=> m MgO = 9,2 - 1,2 = 8 (g)

=> %(m) Mg = (1,2/9,2) . 100%= 13,04 %

%(m) MgO = 100 - 13,04 = 86,96 %