a)

Gọi số mol MgSO₃, MgCO₃ là a, b (mol)

=> 104a + 84b = 1,88 (1)

\(n_{khí}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)

PTHH: MgSO₃ + 2HCl --> MgCl₂ + SO₂ + H₂O

                 a----------------->a-------->a

             MgCO₃ + 2HCl --> MgCl₂ + CO₂  + H₂O

                b------------------>b-------->b

=> a + b = 0,02 (2)

(1)(2) => a = 0,01 (mol); b = 0,01 (mol)

=> \(\left\{{}\begin{matrix}m_{MgSO_3}=0,01.104=1,04\left(g\right)\\m_{MgCO_3}=0,01.84=0,84\left(g\right)\end{matrix}\right.\)

n_MgCl2 = 0,02 (mol)

=> m = 0,02.95 = 1,9 (g)

b)

\(\left\{{}\begin{matrix}n_{SO_2}=0,01\left(mol\right)\\n_{CO_2}=0,01\left(mol\right)\end{matrix}\right.\)

n_KOH = 0,25.0,3 = 0,075 (mol)

Xét tỉ lệ: \(\dfrac{n_{KOH}}{n_{CO_2}+n_{SO_2}}=\dfrac{0,075}{0,01+0,01}=3,75\) => Tạo ra muối K₂CO₃ và K₂SO₃

PTHH: 2KOH + CO₂ --> K₂CO₃ + H₂O

                          0,01---->0,01

            2KOH + SO₂ --> K₂SO₃ + H₂O

                          0,01---->0,01

=> \(\left\{{}\begin{matrix}m_{K_2CO_3}=0,01.138=1,38\left(g\right)\\m_{K_2SO_3}=0,01.158=1,58\left(g\right)\end{matrix}\right.\)

a) Gọi số mol Ca, CaCO₃ là a, b (mol)

=> 40a + 100b = 2,8 (1)

\(n_{khí}=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)

PTHH: Ca + 2HCl --> CaCl₂ + H₂

             a-------------->a------>a

            CaCO₃ + 2HCl --> CaCl₂ + CO₂ + H₂O

               b------------------>b------->b

=> a + b = 0,04 (2)

(1)(2) => a = 0,02 (mol); b = 0,02 (mol)

\(n_{CaCl_2}=a+b=0,04\left(mol\right)\)

=> m = 0,04.111 = 4,44 (g)

\(\left\{{}\begin{matrix}m_{Ca}=0,02.40=0,8\left(g\right)\\m_{CaCO_3}=0,02.100=2\left(g\right)\end{matrix}\right.\)

b) 

\(\overline{M}_X=\dfrac{0,02.2+0,02.44}{0,02+0,02}=23\left(g/mol\right)\)

=> \(d_{X/H_2}=\dfrac{23}{2}=11,5\)

c) 

n_NaOH = 0,1.0,2 = 0,02 (mol)

Xét tỉ lệ: \(\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,02}{0,02}=1\) => Tạo muối NaHCO₃

PTHH: NaOH + CO₂ --> NaHCO₃

              0,02------------>0,02

=> m_NaHCO3 = 0,02.84 = 1,68 (g)

a) \(n_{AlCl_3}=\dfrac{6,675}{133,5}=0,05\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

          0,05<-----------0,05---->0,075

=> \(\%Al=\dfrac{0,05.27}{14,15}.100\%=9,54\%\)

=> \(\%Cu=\dfrac{14,15-0,05.27}{14,15}.100\%=90,46\%\)

b) \(V_{H_2}=0,075.22,4=1,68\left(l\right)\)

c) \(n_{Cu}=\dfrac{14,15-0,05.27}{64}=0,2\left(mol\right)\)

PTHH: 4Al + 3O₂ --t^o--> 2Al₂O₃

          0,05->0,0375

           2Cu + O₂ --t^o--> 2CuO 

            0,2-->0,1

=> \(V_{O_2}=\left(0,1+0,0375\right).22,4=3,08\left(l\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\\ m_{AlCl_3}=6,675\left(mol\right)\\ n_{AlCl_3}=\dfrac{6,675}{133,5}=0,05\left(mol\right)\\ \Rightarrow n_{Al}=n_{AlCl_3}=0,05\left(mol\right)\\ \Rightarrow m_A=0,05.27=1,35\left(g\right);m_{Cu}=14,15-1,35=12,8\left(g\right)\\ \%m_{Cu}=\dfrac{12,8}{14,15}.100\approx90,459\%\\ \Rightarrow\%m_{Al}\approx9,541\%\\ b,n_{Cu}=\dfrac{12,8}{64}=0,2\left(mol\right)\\ n_{H_2}=\dfrac{3}{2}.n_{Al}=\dfrac{3}{2}.0,05=0,075\left(mol\right)\\ \Rightarrow V=V_{H_2\left(đktc\right)}=0,075.22,4=1,68\left(l\right)\\ 4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ 2Cu+O_2\rightarrow\left(t^o\right)2CuO\\ n_{O_2}=\dfrac{3}{4}.n_{Al}+\dfrac{1}{2}.n_{Cu}=\dfrac{3}{4}.0,05+\dfrac{1}{2}.0,2=0,0875\left(mol\right)\)

\(\Rightarrow V_{O_2\left(đktc\right)}=0,0875.22,4=1,96\left(l\right)\)

\(n_{khí}=\dfrac{11.2}{22.4}=0.5\left(mol\right)\)

\(n_{CaCO_3}=a\left(mol\right)\)

\(n_{K_2SO_3}=b\left(mol\right)\)

\(\Rightarrow m_{hh}=100a+158b=70.3\left(g\right)\left(1\right)\)

\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

\(K_2SO_3+2HCl\rightarrow2KCl+SO_2+H_2O\)

\(n_{khí}=a+b=0.5\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.15,b=0.35\)

\(m_{Muối}=m_{CaCl_2}+m_{KCl}=0.15\cdot111+0.35\cdot2\cdot74.5=68.8\left(g\right)\)

Cái chủ đề ko liên quan nha
cám ơn người cmt đã giải bài này
....Thanks you........

CaCO₃+2HCl\(\rightarrow\)CaCl₂+CO₂+H₂O

CaCO₃+H₂SO₄\(\rightarrow\)CaSO₄+CO₂+H₂O

\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25mol\)

Gọi x, y lần lượt là số mol CaCl₂ và CaSO₄.Ta có hệ:

x+y=0,25

111x+136y=32,7

Giải ra x=0,052, y=0,198

Số mol HCl=x=0,052mol

\(C_{M_{HCl}}=\dfrac{0,052}{0,1}=0,52M\)

Số mol H₂SO₄=y=0,198mol

\(C_{M_{H_2SO_4}}=\dfrac{0,198}{0,1}=1,98M\)

\(m_{CaCO_3}=\left(0,052+0,198\right).100=25g\)

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

a)

$RCO_3 + 2HCl \to RCl_2 + CO_2 + H_2O$
$R_2(CO_3)_3 + 6HCl \to 2RCl_3 + 3CO_2 + 3H_2O$

b)

Theo PTHH : 

$n_{H_2O} = n_{CO_2} = \dfrac{0,672}{22,4} = 0,03(mol)$
$n_{HCl} = 2n_{CO_2} = 0,06(mol)$
Bảo toàn khối lượng : 

$m_{muối} = 10 + 0,06.36,5 - 0,03.44 - 0,03.18 = 10,33(gam)$