Mg + 2HCl -> MgCl2 + H2

a                                    a

MgO + 2HCl -> MgCl2 + H2O

 b                                       b

\(nH2=\dfrac{4.48}{22.4}=0.2mol\)\(\Rightarrow a=0.2mol\)

\(\%mMg=\dfrac{0.2\times24\times100}{8.8}=54.5\%\)

\(\%mMgO=100-54.5=45.5\%\)

PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)  (1)

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

a) Ta có: \(n_{H_2}=\dfrac{22,4}{22,4}=1\left(mol\right)=n_{Mg}\) \(\Rightarrow m_{Mg}=1\cdot24=24\left(g\right)\)

\(\Rightarrow\%m_{Mg}=\dfrac{24}{32}\cdot100\%=75\%\) \(\Rightarrow\%m_{MgO}=25\%\)

b) Theo 2 PTHH: \(\left\{{}\begin{matrix}n_{HCl\left(1\right)}=2n_{Mg}=2mol\\n_{HCl\left(2\right)}=2n_{MgO}=2\cdot\dfrac{32-24}{40}=0,4mol\end{matrix}\right.\)

\(\Rightarrow\Sigma n_{HCl}=2,4mol\) \(\Rightarrow m_{ddHCl}=\dfrac{2,4\cdot36,5}{7,3\%}=1200\left(g\right)\)

c) Theo PTHH: \(\Sigma n_{MgCl_2}=\dfrac{1}{2}\Sigma n_{HCl}=1,2mol\)

\(\Rightarrow\Sigma m_{MgCl_2}=1,2\cdot95=114\left(g\right)\)

Mặt khác: \(m_{H_2}=1\cdot2=2\left(g\right)\)

\(\Rightarrow m_{dd}=m_{hh}+m_{ddHCl}-m_{H_2}=1230\left(g\right)\)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{114}{1230}\cdot100\%\approx9,27\%\)

nH2=2.24/22.4=0.1 mol

pt: Mg+2HCL->MgCl2+H2

nMg=nH2=0.1 mol

mMg=0.1*24=2.4g

=>mMg=6-2.4=3.6g

%mMg=(3.6/6)*100=60%

\(n_{H_2}=\dfrac{2.24}{22.4}=0,1\left(mol\right)\)

Ta có phương trình hoá học sau:

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

\(n_{Mg}=n_{H_2}=0,1\left(mol\right)\)

\(m_{Mg}=0,1.24=2,4\left(g\right)\)

\(\Rightarrow m_{Mg}=6-2,4=3,6\left(g\right)\)

\(\Rightarrow\%m_{Mg}=\dfrac{3,6}{6}.100=60\%\)

Chúc bạn học tốt nha!!!

\(n_{H_2}=\dfrac{0,336}{22,4}=0,015(mol)\\ PTHH:Mg+2HCl\to MgCl_2+H_2\\ MgO+2HCl\to MgCl_2+H_2O\\ \Rightarrow n_{Mg}=0,015(mol)\\ \Rightarrow \%_{Mg}=\dfrac{0,015.24}{1,5}.100\%=24\%\\ \Rightarrow \%_{MgO}=100\%-24\%=76\%\)

Chọn A

\(n_{Mg}=n_{H_2}=\dfrac{0.336}{22.4}=0.015\left(mol\right)\)

\(m_{Mg}=0.015\cdot24=0.36\left(g\right)\)

\(\%Mg=\dfrac{0.36}{1.5}\cdot100\%=24\%\)

\(\%MgO=100-24=76\%\)

Cu không pư HCl

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

\(n_{H_2}=\dfrac{6,72}{22,4}= 0,3 mol\)

Theo PTHH:

\(n_{Mg}= n_{H_2}= 0,3 mol\)

\(\Rightarrow m_{Mg}= 0,3 . 24= 7,2g\)

%m_\(Mg\)=\(\dfrac{7,2}{15} . 100\)%= 48%

%m_\(Cu\)= 100% - 48%= 52%

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\)

\(Ca+2HCl\rightarrow CaCl_2+H_2\)

0,1                                  0,1     ( mol )

\(\rightarrow\left\{{}\begin{matrix}\%m_{Ca}=\dfrac{0,1.40}{10}.100=40\%\\\%m_{MgO}=100\%-40\%=60\%\end{matrix}\right.\)

\(\left\{{}\begin{matrix}C\%_{CaCl_2}=\dfrac{0,1.111}{10+390,2-0,1.2}.100=2,775\%\\C\%_{MgO}=\dfrac{4}{10+390,2-0,1.2}.100=1\%\end{matrix}\right.\)

1)

a)

$CaO + 2HCl \to CaCl_2 + H_2O$
$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$

$n_{CaCO_3} = n_{CO_2} = 0,2(mol)$
$n_{CaCl_2} = 0,3(mol)$

Suy ra: 

$n_{CaO} = 0,3 - 0,2 = 0,1(mol)$
$\%m_{CaO} = \dfrac{0,1.56}{0,1.56 + 0,2.100}.100\% = 21,875\%$

$\%m_{CaCO_3} = 78,125\%$

b) 

$m_{dd} = 0,1.56 + 0,2.100 + 50 - 0,2.44 = 66,8(gam)$
$C\%_{CaCl_2} = \dfrac{33,3}{66,8}.100\% = 49,85\%$

Câu 4 : 

a)

Gọi $n_{Fe} = a(mol) ; n_{MgO} = b(mol)$

Suy ra: $56a + 40b = 19,2(1)$

$Fe + 2HCl \to FeCl_2 + H_2$
$MgO + 2HCl \to MgCl_2 + H_2O$
Theo PTHH : $n_{HCl} = 2a + 2b = 0,4.2 = 0,8(2)$
Từ (1)(2) suy ra a = b = 0,2

$\%m_{Fe} = \dfrac{0,2.56}{19,2}.100\% = 58,33\%$
$\%m_{MgO} = 100\% -58,33\% = 41,67\%$

b)

$n_{FeCl_2} = a = 0,2(mol)$
$n_{MgCl_2} = b = 0,2(mol)$

$m_{muối} = 0,2.127 + 0,2.95 = 44,4(gam)$

Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\n_{CaCl_2}=\dfrac{33,3}{111}=0,3\left(mol\right)\end{matrix}\right.\)

PTHH: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2\uparrow+H_2O\)

                    0,2_____0,4_____0,2____0,2_____0,2  (mol)

            \(CaO+2HCl\rightarrow CaCl_2+H_2O\)

                 0,1_____0,2_____0,1____0,1    (mol)

Ta có: \(\left\{{}\begin{matrix}m_{CaO}=0,1\cdot56=5,6\left(g\right)\\m_{CaCO_3}=0,2\cdot100=20\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{CaO}=\dfrac{5,6}{5,6+20}\cdot100\%=21,875\%\\\%m_{CaCO_3}=78,125\%\end{matrix}\right.\)

Mặt khác: \(m_{CO_2}=0,2\cdot44=8,8\left(g\right)\)

\(\Rightarrow m_{dd\left(sau.p/ứ\right)}=m_{CaO}+m_{CaCO_3}+m_{ddHCl}-m_{CO_2}=66,8\left(g\right)\)

\(\Rightarrow C\%_{CaCl_2}=\dfrac{33,3}{66,8}\cdot100\%\approx49,85\%\)

nCO2=\(\dfrac{4,48}{22,4}=0,2\) mol

nCaCl₂=\(\dfrac{33,3}{111}=0,3\)

CaCO₂ + 2HCl → CaCl₂ + CO₂ + H₂O

  0,2           ←         0,2    ← 0,2

CaO + 2HCl  → CaCl₂ + H₂O     

 0,1              ←    0,1

a)  % CaO=\(\dfrac{0,1.56}{0,1.56+0,2.100}.100\%=21,875\%\)    

   % CaCO₃ =100% - 21,875%= 78,125%

b) a = mCaO+mCaCO₃ =0,1.56+0,2.100=25,6g

mdd sau pư= a + mddHCl - mCO2

                  = 25,6 + 50 - 0,2.44=66,8g

C%CaCl₂=\(\dfrac{33,3}{66,8}.100\%\simeq49,85\%\)