a, \(n_{C_2H_4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

PT: \(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)

\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)

Theo PT: \(n_{CaCO_3}=n_{CO_2}=2n_{C_2H_4}=0,1\left(mol\right)\)

\(\Rightarrow m_{CaCO_3}=0,1.100=10\left(g\right)\)

b, Theo PT: \(n_{O_2}=3n_{C_2H_4}=0,15\left(mol\right)\Rightarrow V_{O_2}=0,15.22,4=3,36\left(l\right)\)

\(\Rightarrow V_{kk}=\dfrac{V_{O_2}}{20\%}=16,8\left(l\right)\)

a, \(n_{CH_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH:

CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

0,5--->1------------->0,5

Ca(OH)₂ + CO₂ ---> CaCO₃ + H₂O

                   0,5----->0,5

b, \(V_{O_2}=1.22,4=22,4\left(l\right)\)

c, \(m_{CaCO_3}=0,5.100=50\left(g\right)\)

A tác dụng với NaHCO₃ cho khí CO2 → A: axit CH₃COOH

BTKL: m + m_O2 = m_CO2 + m_H2O => m = 1,8

=> n_CH3COOH = 0,03

CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O

0,03                 0,02                 0,0125

=> H = 62,5%

\(n_{C_2H_5OH}=\dfrac{9,2}{46}=0,2\left(mol\right)\)

\(C_2H_5OH+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)

0,2              0,6         0,4      0,4 

\(a,V_{O_2}=0,6.22,4=13,44\left(l\right)\)

\(V_{kk}=13,44.5=67,2\left(l\right)\)

b, \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\)

    0,4                               0,4 

\(m_{CaCO_3}=0,4.100=40\left(g\right)\)

\(m_{CaCO_3tt}=40.95\%=38\left(g\right)\)

a) \(n_{C_2H_4}=\dfrac{4,958}{24,79}=0,2\left(mol\right)\)

PTHH: \(C_2H_4+3O_2\xrightarrow[]{t^o}2CO_2+2H_2O\)

              0,2---->0,6---->0,4---->0,4

\(\Rightarrow V_{O_2}=0,6.24,79=14,874\left(l\right)\)

b) \(V_{CO_2}=0,4.22,4=9,916\left(l\right)\)

c) \(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3\downarrow+H_2O\)

                       0,4------>0,4

\(\Rightarrow\left\{{}\begin{matrix}x=m_{CaCO_3}=0,4.100=40\left(g\right)\\y=m_{b\text{ình}.t\text{ăng}}=m_{CO_2}+m_{H_2O}=0,4.44+0,4.18=24,8\left(g\right)\end{matrix}\right.\)

PTHH: \(C_2H_5OH+3O_2\xrightarrow[]{t^o}2CO_2+3H_2O_{ }\)

           \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\)

a) Ta có: \(n_{CaCO_3}=\dfrac{160}{100}=1,6\left(mol\right)=n_{CO_2}\) \(\Rightarrow n_{O_2}=2,4\left(mol\right)\)

\(\Rightarrow V_{kk}=\dfrac{2,4\cdot22,4}{20\%}=268,8\left(l\right)\)

b) Theo PTHH: \(n_{C_2H_5OH}=\dfrac{1}{2}n_{CO_2}=0,8\left(mol\right)\)

\(\Rightarrow a=C\%_{C_2H_5OH}=\dfrac{0,8\cdot46}{50\cdot0,8}\cdot100\%=92\%=92^o\)

nCH4 = 11.2/22.4 = 0.5 (mol) 

CH4 + 2O2 -to-> CO2 + 2H2O 

0.5____________0.5 

CO2 + Ca(OH)2 => CaCO3 + H2O 

0.5_______________0.5 

mCaCO3 = 0.5*100 = 50 (g) 

\(CH_4 + 2O_2 \xrightarrow{t^o} CO_2 +2H_2O\\ CO_2 +C a(OH)_2 \to CaCO_3 + H_2O\\ n_{CaCO_3} = n_{CO_2} = n_{CH_4} = \dfrac{11,2}{22,4} = 0,5(mol)\\ m = 0,5.100 = 50(gam)\)