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

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

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

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%

$n_{Al_2O_3} = \dfrac{4,08}{102} = 0,04(mol)$

\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

0,04          0,12                    0,04                                (mol)

$m_{Al_2(SO_4)_3} = 0,04.342 = 13,68(gam)$

$C\%_{H_2SO_4} = \dfrac{0,12.98}{280}.100\% = 4,2\%$

\(m_{CH_3COOH}=150.12\%=18g\)

\(n_{CH_3COOH}=\dfrac{18}{60}=0,3mol\)

\(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)

      0,3                   0,15                0,3                 0,15               ( mol )

\(m_{ddNa_2CO_3}=\left(0,15.106\right):10,6\%=150g\)

\(V_{CO_2}=0,15.22,4=3,36l\)

\(m_{CH_3COONa}=0,3.82=24,6g\)

\(m_{ddspứ}=150+150-0,15.44=293,4g\)

\(C\%_{CH_3COONa}=\dfrac{24,6}{293,4}.100=8,28\%\)

PTHH: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

Ta có: \(n_{Fe_2O_3}=\dfrac{4,8}{160}=0,03\left(mol\right)\) 

\(\Rightarrow\left\{{}\begin{matrix}n_{H_2SO_4}=0,09\left(mol\right)\\n_{Fe_2\left(SO_4\right)_3}=0,03\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddH_2SO_4}=\dfrac{0,09\cdot98}{9,8\%}=90\left(g\right)\\m_{Fe_2\left(SO_4\right)_3}=0,03\cdot400=12\left(g\right)\end{matrix}\right.\)

\(a.Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\\ n_{NaCl}=n_{HCl}=2.n_{CO_2}=2.\dfrac{448:1000}{22,4}=0,04\left(mol\right)\\ C_{MddHCl}=\dfrac{0,04}{0,02}=2\left(M\right)\\ b.m_{NaCl}=58,5.0,04=2,34\left(g\right)\\ c.m_{Na_2CO_3}=106.0,02=2,12\left(g\right)\\ \%m_{Na_2CO_3}=\dfrac{2,12}{5}.100=42,4\%\\ \%m_{NaCl}=100\%-42,4\%=57,6\%\)

Bài 16 : 

\(n_{CO2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)

Pt : \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O|\)

            1              2                2           1         1

           0,02         0,04           0,04     0,02

a) \(n_{HCl}=\dfrac{0,02.2}{1}=0,04\left(mol\right)\)

 20ml = 0,02l

\(C_{M_{HCl}}=\dfrac{0,04}{0,02}=2\left(M\right)\)

b) \(n_{NaCl}=\dfrac{0,02.2}{1}=0,04\left(mol\right)\)

⇒ \(m_{NaCl}=0,04.58,5=2,34\left(g\right)\)

c) \(n_{Na2CO3}=\dfrac{0,04.1}{2}=0,02\left(mol\right)\)

 ⇒ \(m_{Na2CO3}=0,02.106=2,12\left(g\right)\)

\(m_{NaCl}=5-2,12=2,88\left(g\right)\)

⁰/₀Na₂CO₃ = \(\dfrac{2,12.100}{5}=42,4\)⁰/₀

⁰/₀NaCl = \(\dfrac{2,88.100}{5}=57,6\)⁰/₀

 Chúc bạn học tốt

\(n_{Al}=\dfrac{21,6}{27}=0,8\left(mol\right)\)

Pt : \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O|\)

        2             3                  1                3

       0,8          1,2               0,4             1,2

a) \(n_{H2}=\dfrac{0,8.3}{2}=1,2\left(mol\right)\)

\(V_{H2\left(dktc\right)}=1,2.22,4=26,88\left(l\right)\)

b) \(n_{H2SO4}=\dfrac{0,8.3}{2}=1,2\left(mol\right)\)

⇒ \(m_{H2SO4}=1,2.98=117,6\left(g\right)\)

\(m_{ddH2SO4}=\dfrac{117,6.100}{29,4}=400\left(g\right)\)

c) \(n_{Al2\left(SO4\right)3}=\dfrac{1,2.1}{3}=0,4\left(mol\right)\)

⇒ \(m_{Al2\left(SO4\right)3}=0,4.342=136,8\left(g\right)\)

\(m_{ddspu}=21,6+400-\left(1,2.2\right)=419,2\left(g\right)\)

\(C_{Al2\left(SO4\right)3}=\dfrac{136,8.100}{419,2}=32,63\)⁰/₀

 Chúc bạn học tốt

\(n_{CuO}=\dfrac{1.6}{80}=0.02\left(mol\right)\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(0.02.......0.02.................0.02\)

\(m_{H_2SO_4}=0.02\cdot98=1.96\left(g\right)\)

\(m_{dd_{H_2SO_4}}=\dfrac{1.96}{20\%}=9.8\left(g\right)\)

\(m_{\text{dung dịch sau phản ứng }}=1.6+9.8=11.4\left(g\right)\)

\(C\%_{CuSO_4}=\dfrac{0.02\cdot160}{11.4}=28.07\%\)