-nNa₂CO₃= m/M = 10,6/106 = 0,1 (mol)

-PT:Na₂CO₃+CaCl₂->CaCO₃+2NaCl

____0,1____________0,1______0,2

-mCaCO₃= n.M = 0,1.100 = 10 (g)

-mNaCl= n.M = 0,2.58,5 = 11,7 (g)

\(n_{Na_2CO_3}=\dfrac{10.6}{106}=0.1\left(mol\right)\\ PTHH:CaCl_2+Na_2CO_3\rightarrow CaCO_3+2NaCl\\ n_{CaCO_3}=n_{Na_2CO_3}=0,1\left(mol\right)\\ \Rightarrow m_{CaCO_3}=0,1.100=10g\)

$n_{NaCl}=n_{Na_2CO_3}=0,1.2=0,2(mol)$

`=>` $m_{NaCl}=0.2.58,5=11,7g$

n_Na2CO3 = 0,1 mol

Na₂CO₃ + CaCl₂ → CaCO₃ + 2NaCl

0,1..............0,1................0,1............0,2

⇒ m_CaCO3 = 0,1.100 = 10 (g)

⇒ m_NaCl = 0,2.58,5 = 11,7 (g)

8tk

\(n_{Na2CO3}=0,1mol\)

\(Na_2CO_3+CaCL_2\rightarrow CaCO_3+2NaCL\)

\(0,1..........0,1..........0,1..........0,2..........\)

\(\Rightarrow m_{CaCO3}=0,1\times100=10g\)

\(\Rightarrow m_{NaCL}=0,2\times58,5=11,7\left(g\right)\)

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Chúc bạn học tốt !!!

Ê ! On mà k tl t hả thk kia ( ngoài lề ) 

K liên wan tới mí bợn ạ .

\(n_{H_2SO_4}=\dfrac{150.9,8\%}{98}=0,15\left(mol\right)\\ H_2SO_4+Na_2CO_3\rightarrow Na_2SO_4+H_2O+CO_2\\ n_{Na_2CO_3}=n_{H_2SO_4}=0,15\left(mol\right)\\ \Rightarrow m_{ddNa_2CO_3}=\dfrac{0,15.106}{10,6\%}=150\left(g\right)\\ n_{CO_2}=n_{H_2SO_4}=0,15\left(mol\right)\\ m_{ddsaupu}=150+150-0,15.44=293,4\left(g\right)\\ n_{Na_2SO_4}=n_{H_2SO_4}=0,15\left(mol\right)\\ C\%_{Na_2SO_4}=\dfrac{0,15.142}{293,4}.100=7,26\%\)

\(PTHH:CaCl_2+2AgNO_3\rightarrow Ca\left(NO_3\right)_2+2AgCl\downarrow\)

\(n_{CaCl_2}=\frac{22,2}{111}=0,2\left(mol\right)\)

Theo PT: \(n_{AgNO_3}=2n_{CaCl_2}=0,4\left(mol\right)\)

\(\Rightarrow m_{AgNO_3}=0,4.170=68\left(g\right)\)

b) Các chất còn lại trong phản ứng là Ca(NO₃)₂, AgCl

\(TheoPT:n_{Ca\left(NO_3\right)_2}=n_{CaCl_2}=0,2\left(mol\right)\)

\(n_{AgCl}=2n_{CaCl_2}=0,4\left(mol\right)\)

\(\Rightarrow m_{Ca\left(NO_3\right)_2}=0,2.164=32,8\left(g\right)\)

\(m_{AgCl}=0,4.143,5=57,4\left(g\right)\)

CaCl₂ + 2AgNO₃ → Ca(NO₃)₂ + 2AgCl

\(n_{CaCl_2}=\frac{22,2}{111}=0,2\left(mol\right)\)

a) Theo PT: \(n_{AgNO_3}=2n_{CaCl_2}=2\times0,2=0,4\left(mol\right)\)

\(\Rightarrow m_{AgNO_3}=0,4\times170=68\left(g\right)\)

b) Theo PT: \(n_{Ca\left(NO_3\right)_2}=n_{CaCl_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Ca\left(NO_3\right)_2}=0,2\times164=32,8\left(g\right)\)

Theo PT: \(n_{AgCl}=n_{AgNO_3}=0,4\left(mol\right)\)

\(\Rightarrow m_{AgCl}=0,4\times143,5=57,4\left(g\right)\)

Na₂CO₃ + CaCl₂ → 2NaCl + CaCO₃

\(n_{Na_2CO_3}=\frac{10,6}{106}=0,1\left(mol\right)\)

Theo pT: \(n_{CaCO_3}=n_{Na_2CO_3}=0,1\left(mol\right)\)

\(\Rightarrow m_{CaCO_3}=0,1\times100=10\left(g\right)\)

Theo pT: \(n_{NaCl}=2n_{Na_2CO_3}=2\times0,1=0,2\left(mol\right)\)

\(\Rightarrow m_{NaCl}=0,2\times58,5=11,7\left(g\right)\)

\(n_{H_2SO_4}=0.25\cdot2=0.5\left(mol\right)\)

\(Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+CO_2+H_2O\)

\(0.5..............0.5...............0.5\)

\(m_{Na_2CO_3}=0.5\cdot106=53\left(g\right)\)

\(C_{M_{Na_2SO_4}}=\dfrac{0.5}{0.25}=2\left(M\right)\)

\(\)250ml = 0,25l

\(n_{H2SO4}=2.0,25=0,5\left(mol\right)\)

Pt : \(Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+CO_2+H_2O|\)

            1                1                 1            1          1

          0,5               0,5              0,5

\(n_{Na2CO3}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)

⇒ \(m_{Na2CO3}=0,5.106=53\left(g\right)\)      

\(n_{Na2SO4}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)

\(C_{M_{Na2SO4}}=\dfrac{0,5}{0,25}=2\left(M\right)\)

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\(m_{CH_3COOH}=24\%.150=36\left(g\right)\\ \rightarrow n_{CH_3COOH}=\dfrac{36}{60}=0,6\left(mol\right)\)

PTHH: 2CH₃COOH + Na₂CO₃ ---> 2CH₃COONa + CO₂ + H₂O

                0,6                  0,3                  0,6                  0,3 

=> V_CO2 = 0,3.22,4 = 6,72 (l)

\(m_{Na_2CO_3}=0,3.31,8\left(g\right)\)

=> \(m_{ddNa_2CO_3}=\dfrac{31,8}{21,2\%}=150\left(g\right)\)

m_CO2 = 0,3.44 = 13,2 (g)

\(m_{dd}=150+150-13,2=286,8\left(g\right)\)

\(m_{CH_3COONa}=0,3.82=24,6\left(g\right)\\ \rightarrow C\%_{CH_3COONa}=\dfrac{24,6}{286,8}=8,58\%\)

\(n_{CaCl_2}=\dfrac{22.2}{111}=0.2\left(mol\right)\)

\(CaCl_2+2AgNO_3\rightarrow Ca\left(NO_3\right)_2+2AgCl\)

\(0.2....................................................0.4\)

\(m_{AgCl}=0.4\cdot143.5=57.4\left(g\right)\)

n_CaCl2=22,2/111=0,2(mol)

CaCl₂ + 2AgNO₃ -----> 2AgCl + Ca(NO₃)₂

TPT:n_AgCl=2.n_CaCl2=2.0,2=0,4(mol)

m_AgCl=0,4.143,5=57,4(g)

\(m_{H_2SO_4}=150.9,8\%=14,7\left(g\right)\\ n_{H_2SO_4}=\dfrac{14,7}{98}=0,3\left(mol\right)\\ PTHH:H_2SO_4+Na_2CO_3\rightarrow Na_2SO_4+CO_2\uparrow+H_2O\\ Mol:0,3\rightarrow0,3\rightarrow0,3\rightarrow0,3\)

\(m_{Na_2CO_3}=0,3.106=31,8\left(g\right)\\ m_{ddNa_2CO_3}=\dfrac{31,8}{10,6\%}=300\left(g\right)\\ m_{Na_2SO_4}=0,3.142=42,6\left(g\right)\\ m_{CO_2}=0,3.44=13,2\left(g\right)\\ m_{dd}=150+300-13,2=436,8\left(g\right)\\ C\%_{Na_2SO_4}=\dfrac{42,6}{436,8}=9,75\%\)

mH2SO4 =mdd H2SO4.C% : 100% = 400.9,8% :100% = 39,2 (g)

=> nH2SO4 = mH2SO4 : MH2SO4 = 39,2: 98 = 0,4 (mol)

PTHH: H2SO4 + Na2CO3 ---> Na2SO4 + CO2 + H2O

0,4 ---->0,4 -----------> 0,4 -------> 0,4 (mol)

a) Theo PTHH: nNa2CO3 = nH2SO4 = 0,4 (mol)

=> mNa2CO3 = nNa2CO3. MNa2CO3 = 0,4.106 = 42,4 (g)

=> mdd Na2CO3 = mNa2CO3. 100% : C% = 42,4.100% : 10% = 424 (g)

b) Theo PTHH: nCO2 = nH2SO4 = 0,4 (mol)

=> VCO2(đktc) = 0,4.22,4 = 8,96 (lít)

c) Theo PTHH: nNa2SO4 = nH2SO4 = 0,4 (mol)

=> mNa2SO4 = nNa2SO4. MNa2SO4 = 0,4.142 = 56,8 (g)

mdd A = mdd H2SO4 + mdd Na2CO3 = 400 + 424 = 824 (g)

dd A chứa Na2SO4

=> C% Na2SO4 = (mNa2SO4 : mddA).100% = (56,8 : 824).100% = 6,89%

a) 

\(n_{Na_2CO_3}=\dfrac{10,6}{106}=0,1\left(mol\right)\)

PTHH: Na₂CO₃ + 2CH₃COOH --> 2CH₃COONa + CO₂ + H₂O

                0,1--------->0,2------------->0,2------------>0,1

=> m_CH3COOH = 0,2.60 = 12 (g)

\(C_{M\left(dd.CH_3COOH\right)}=\dfrac{0,2}{0,4}=0,5M\)

b) \(n_{C_2H_5OH}=\dfrac{13,8}{46}=0,3\left(mol\right)\)

PTHH: CH₃COOH + C₂H₅OH --H₂SO₄,t^o--> CH₃COOC₂H₅ + H₂O

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,3}{1}\) => Hiệu suất tính theo CH₃COOH

\(n_{CH_3COOH\left(pư\right)}=\dfrac{0,2.80}{100}=0,16\left(mol\right)\)

PTHH: CH₃COOH + C₂H₅OH --H₂SO₄,t^o--> CH₃COOC₂H₅ + H₂O

                 0,16------------------------------------->0,16

=> \(m_{CH_3COOC_2H_5}=0,16.88=14,08\left(g\right)\)