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

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

        1          2               1         1

       0,05    0,1           0,05      0,05 

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

       1           2              1            1

      0,2       0,4            0,2

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

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

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

⁰/₀Mg = \(\dfrac{1,2.100}{9,2}=13,04\)⁰/₀

⁰/₀MgO = \(\dfrac{8.100}{9,2}=86,96\)⁰/₀

b) Có : \(m_{MgO}=8\left(g\right)\)

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

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

\(m_{HCl}=0,5.36,5=18,25\left(g\right)\)

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

c) \(n_{MgCl2\left(tổng\right)}=0,05+0,2=0,25\left(mol\right)\)

⇒ \(m_{MgCl2}=0,25.95=23,75\left(g\right)\)

\(m_{ddspu}=9,2+125-\left(0,05.2\right)=134,1\left(g\right)\)

\(C_{MgCl2}=\dfrac{23,75.100}{134,1}=17,71\)⁰/₀

 Chúc bạn học tốt

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

PTHH: Mg + 2HCl → MgCl₂ + H₂

Mol:    0,05    0,1           0,05        0,05

PTHH: MgO + 2HCl → MgCl₂ + H₂O

Mol:     0,2         0,4         0,2

\(\%m_{Mg}=\dfrac{0,05.24.100\%}{9,2}=13,04\%;\%m_{MgO}=100-13,04=86,96\%\)

\(n_{MgO}=\dfrac{9,2-0,05.24}{40}=0,2\left(mol\right)\)

b,\(m_{HCl}=\left(0,1+0,4\right).36,5=18,25\left(g\right)\)

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

c,m_{dd sau pứ} = 9,2+125-0,05.2 = 134,1 (g)

 \(C\%_{ddMgCl_2}=\dfrac{\left(0,05+0,2\right).95.100\%}{134,1}=17,71\%\)

\(Mg + 2HCl \rightarrow MgCl_2 + H_2\)     (1)

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

Khí thu được là H₂

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

Theo PTHH (1):

\(n_{Mg}= n_{H_2}= 0,05 mol\)

\(\Rightarrow m_{Mg}= 0,05 . 24= 1,2 g\)

\(\Rightarrow m_{MgO}= 9,2 - 1,2= 8g\)

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

C%_\(MgO\)= 100% - 13,04%=86,96%

b)

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

Theo PTHH (1) và (2):

\(n_{HCl(1)}= 2n_{Mg}= 0,1 mol\)

\(n_{HCl(2)}= 2n_{MgO}= 0,4 mol\)

Suy ra: \(n_{HCl}= n_{HCl(1)} + n_{HCl(2)}\)

                     \(= 0,1 + 0,4= 0,5 mol\)

\(\Rightarrow m_{HCl}= 0,5 . 36,5= 18,25g\)

\(\Rightarrow m_{dd HCl} = \dfrac{18,25 . 100%}{14,6%}=125 g\)

_c)

\(\)Dung dịch sau pư: MgCl₂

Theo PTHH:

\(n_{MgCl_2}= \dfrac{1}{2} n_{HCl}= 0,25 mol\)

\(\Rightarrow m_{MgCl_2}= 0,25 . 95=23,75g\)

\(m_{dd sau pư} = m_{Mg} + m_{MgO} + m_{dd HCl}- m_{H_2}\)

               \(= 9,2 + 125 + 2 . 0,05\)

             \(=134,1 g\)

C%_{\(MgCl_2\)}=\(\dfrac{23,75}{134,1}. 100\)%=17,71%

 PTHH

Mg + 2HCl ----> MgCl2 + H2      (1)

MgO + 2HCl -----> MgCl2 + H2O (2)

a) Theo pt(1) n Mg = n H2 = \(\frac{1,12}{22,4}\) = 0,05 (mol)

==> m Mg = 0,005 . 24=1,2 (g)

%m Mg =  \(\frac{1,2}{3,2}\). 100%= 37,5%

%m MgO= 100% - 37,5%= 62,5%

b)m dd sau pư = 3,2 + 246,9 - 0,05 . 2=250 (g)

Theo pt(1)(2)  n MgCl2(1) = n Mg = 0,05 mol

                         n MgCl2 (2) = n MgO=\(\frac{3,2-1,2}{40}\)=0,05(mol)

==> tổng n MgCl2 = 0,1 (mol)  ---->m MgCl2 = 9,5 (g)

C%(MgCl2)= \(\frac{9,5}{250}\) .100% = 3,8%

Gọi : \(\left\{{}\begin{matrix}n_{Al_2O_3}=a\left(mol\right)\\n_{Zn}=b\left(mol\right)\end{matrix}\right.\)⇒ 102a + 65b = 2,505(1)

\(Al_2O_3 + 6HCl \to 2AlCl_3 + 3H_2O\\ Zn + 2HCl \to ZnCl_2 + H_2\)

Muối gồm : \(\left\{{}\begin{matrix}AlCl_3:2a\left(mol\right)\\ZnCl_2:b\left(mol\right)\end{matrix}\right.\)⇒ 133,5.2a + 136b = 6,045(2)

Từ (1)(2) suy ra : a = 0,015 ; b = 0,015

Vậy :

\(\%m_{Al_2O_3} = \dfrac{0,015.102}{2,505}.100\% = 61,08\%\\ \%m_{Zn} = 100\% - 61,08\% = 38,92\%\)

Theo PTHH : \(n_{HCl} = 6a + 2b = 0,12(mol)\\ \Rightarrow C\%_{HCl} = \dfrac{0,12.36,5}{200}.100\% = 2,19\%\)

a)

Gọi $n_{Mg} = a ; n_{Al} = b \Rightarrow 24a + 27b = 5,1(1)$
$Mg + 2HCl \to MgCl_2  + H_2$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$

Ta có :

$n_{H_2} = a + 1,5b = \dfrac{5,6}{22,4} = 0,25(2)$
Từ (1)(2) suy ra a = b = 0,1

$\%m_{Mg} = \dfrac{0,1.24}{5,1}.100\%  =47,06\%$

$\%m_{Al} = 52,94\%$

b)

$n_{HCl} = 2n_{H_2} = 0,5(mol)$
$m_{dd\ HCl} = \dfrac{0,5.36,5}{10\%} = 182,5(gam)$

c)

$MgCl_2 + 2NaOH \to Mg(OH)_2 + 2NaCl$
$AlCl_3 + 3NaOH \to Al(OH)_3 + 3NaCl$

$Al(OH)_3 + NaOH \to NaAlO_2 + 2H_2O$

$n_{Mg(OH)_2} = a = 0,1(mol)$
$\Rightarrow m_{kết\ tủa}  = 0,1.58 = 5,8(gam)$

Ta có:

\(Mg+2HCl\rightarrow MgCl_2+H_2\) ; \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

Đặt số mol Mg và Al lần lượt là a và b (a,b>0)
theo bài ra ta có hệ 

\(\left\{{}\begin{matrix}24a+27b=5,1\\a+1,5b=\dfrac{5,6}{22,4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\)

\(\Rightarrow\%Mg=\dfrac{0,1\times24}{5,1}=47,06\%\Rightarrow\%Al=100\%-47,06\%=52,94\%\)

Theo PT có \(n_{HCl}=2n_{Mg}+3n_{Al}=2\times0,1+3\times0,1=0,5\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,5\times36,5=18,25\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{18,25}{10\%}=182,5\left(g\right)\)

\(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2\downarrow+2NaCl\)

\(AlCl_3+3NaOH\rightarrow Al\left(OH\right)_3\downarrow+3NaCl\)

+ Với NaOH vừa đủ

\(a=m_{Mg\left(OH\right)_2}+m_{Al\left(OH\right)_3}=0,1\times58+0,1\times78=13,6\left(g\right)\)

+ Với NaOH dư có thêm PT

\(Al\left(OH\right)_3+NaOH\rightarrow NaAlO_2+2H_2O\)

\(\Rightarrow a=m_{Mg\left(OH\right)_2}=0,1\times58=5,8\left(g\right)\)

Fe₂O₃ + 6HCl → 2FeCl₃ + 3H₂O (1)

CuO + 2HCl → CuCl₂ + H₂O (2)

a) \(m_{CuO}=20\times20\%=4\left(g\right)\)

\(\Rightarrow m_{Fe_2O_3}=20-4=16\left(g\right)\)

b) \(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

Theo PT1: \(n_{HCl}=6n_{Fe_2O_3}=6\times0,1=0,6\left(mol\right)\)

Theo PT2: \(n_{HCl}=2n_{CuO}=2\times0,05=0,1\left(mol\right)\)

\(\Rightarrow\Sigma n_{HCl}=0,1+0,6=0,7\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,7\times36,5=25,55\left(g\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{25,55}{5,475\%}=466,67\left(g\right)\)

c) Dung dịch sau phản ứng gồm: CuCl₂ và FeCl₃

Theo PT1: \(n_{FeCl_3}=2n_{Fe_2O_3}=2\times0,1=0,2\left(mol\right)\)

\(\Rightarrow m_{FeCl_3}=0,2\times162,5=32,5\left(g\right)\)

Theo PT2: \(n_{CuCl_2}=n_{CuO}=0,05\left(mol\right)\)

\(\Rightarrow m_{CuCl_2}=0,05\times135=6,75\left(g\right)\)

\(\Sigma m_{dd}=20+466,67=486,67\left(g\right)\)
\(\Rightarrow C\%_{FeCl_3}=\dfrac{32,5}{486,67}\times100\%=6,68\%\)

\(C\%_{CuCl_2}=\dfrac{6,75}{486,67}\times100\%=1,39\%\)