a) \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

PTHH: Fe + 2HCl → FeCl₂ + H₂

Mol:     0,2     0,4         0,2      0,2

b) \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

c) \(C\%_{ddHCl}=\dfrac{0,4.36,5.100\%}{300}=4,87\%\)

d) m_{dd sau pứ} = 11,2 + 300 - 0,2.2 = 310,8 (g)

\(C\%_{ddFeCl_2}=\dfrac{0,2.127.100\%}{310,5}=8,17\%\)

\(n_{Na_2CO_3}=0,1.1=0,1\left(mol\right)\)

a. \(Na_2CO_3+Ba\left(OH\right)_2\rightarrow BaCO_3+2NaOH\)

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

b. \(m_{kt}=m_{BaCO_3}=0,1.197=19,7\left(g\right)\)

c. \(C\%_{Ba\left(OH\right)_2}=\dfrac{0,1.171.100}{200}=8,55\%\)

d. \(BaCO_3+2HCl\rightarrow BaCl_2+H_2O+CO_2\)

      0,1           0,2

=> \(a=m_{dd.HCl}=\dfrac{0,2.36,5.100}{30}=\dfrac{73}{3}\left(g\right)\)

\(a,PTHH:Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\uparrow\\ b,n_{Na_2CO_3}=\dfrac{15,9}{106}=0,15\left(mol\right)\\ \Rightarrow n_{HCl}=0,3\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,3\cdot36,5=10,95\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{10,95}{200}\cdot100\%=5,475\%\\ c,n_{CO_2}=0,15\left(mol\right)\\ \Rightarrow V_{CO_2\left(đkc\right)}=0,15\cdot24,79=3,7185\left(l\right)\\ d,m_{CO_2}=0,15\cdot44=6,6\left(g\right)\\ n_{NaCl}=0,3\left(mol\right);n_{H_2O}=0,15\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}m_{CT_{NaCl}}=0,3\cdot58,5=17,55\left(g\right)\\m_{H_2O}=0,15\cdot18=2,7\left(g\right)\end{matrix}\right.\\ m_{dd_{NaCl}}=15,9+200-2,7-6,6=206,6\left(g\right)\\ \Rightarrow C\%_{NaCl}=\dfrac{17,55}{206,6}\cdot100\%\approx8,49\%\)

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

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)

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

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

a/ 2NaOH + CuSO₄ -----> Na₂SO₄ + Cu(OH)₂

b/ \(m_{CuSO_4}=200.b\%\) \(\Rightarrow n_{CuSO_4}=\frac{200.b\%}{160}=\frac{5.b\%}{4}\) (mol)

\(m_{NaOH}=4\%.150=6\left(g\right)\) \(\Rightarrow n_{NaOH}=\frac{6}{40}=0,15\left(mol\right)\)

Theo đề bài thì 2nNaOH = nCuSO4

\(\Rightarrow\frac{5}{4}.b\%=0,3\Rightarrow b\%=0,24\%\)

gấp á mn giúp em với ;-;

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

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

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

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

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

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

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

\(m_{ddspứ}=150+150-0,15.44=293,4g\)

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