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

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

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 

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

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

Câu 1.

\(V_{C_2H_5OH}=\dfrac{90.90}{100}=81\left(ml\right)\)

\(m_{C_2H_5OH}=81.0,8=64,8g\)

\(n_{C_2H_5OH}=\dfrac{64,8}{46}=1,4mol\)

\(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\)

    1,4                                                   0,7        ( mol )

\(V_{H_2}=0,7.22,4=15,68l\)

Câu 2.

\(n_{CaCO_3}=\dfrac{100}{100}=1mol\)

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

                        1            1                      ( mol )

\(C_2H_5OH+3O_2\rightarrow\left(t^o\right)2CO_2+3H_2O\)

       0,5          1,5             1                          ( mol )

\(V_{kk}=\left(1,5.22,4\right).5=168l\)           

\(m_{C_2H_5OH}=0,5.46=23g\)  

\(V_{C_2H_5OH}=\dfrac{23}{0,8}=28,75ml\)

Độ rượu = \(\dfrac{28,75}{30}.100=95,83^o\)

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%

\(n_{CH_4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

             0,2      0,4             0,2

CO₂ + Ca(OH)₂ ---> CaCO₃ + H₂O

0,2                               0,2 

\(\rightarrow\left\{{}\begin{matrix}x=0,2.100=20\left(g\right)\\V_{kk}=0,4.5.22,4=44,8\left(l\right)\end{matrix}\right.\)

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

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

\(M_X=\dfrac{1.5}{\dfrac{1.775}{71}}=60\left(g\text{/}mol\right)\)

\(n_X=\dfrac{2.25}{60}=0.0375\left(mol\right)\)

CT : \(C_xH_yO_z\)

\(n_{CaCO_3}=n_{CO_2}=\dfrac{11.25}{100}=0.1125\left(mol\right)\)

\(\Rightarrow n_C=0.1125\left(mol\right)\)

\(m_{bt}=m_{CO_2}+m_{H_2O}=7.65\left(g\right)\)

\(\Rightarrow n_{H_2O}=\dfrac{7.65-0.1125\cdot44}{18}=0.15\left(mol\right)\)

\(\Rightarrow n_H=0.15\cdot2=0.3\left(mol\right)\)

Ta có : 

\(x=\dfrac{n_C}{n_X}=\dfrac{0.1125}{0.0375}=3\)

\(y=\dfrac{n_H}{n_X}=\dfrac{0.3}{0.0375}=8\)

\(M=12\cdot3+8+16z=60\left(g\text{/}mol\right)\)

\(\Rightarrow z=1\)

\(CT:C_3H_8O\)

a, \(C_2H_6O+3O_2\underrightarrow{t^o}2CO_2+3H_2O\)

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

Ta có: \(n_{CaCO_3}=\dfrac{14,4}{100}=0,144\left(mol\right)\)

Theo PT: \(n_{CO_2}=n_{CaCO_3}=0,144\left(mol\right)\Rightarrow m_{CO_2}=0,144.44=6,336\left(g\right)\)

b, \(n_{C_2H_6O}=\dfrac{1}{2}n_{CO_2}=0,072\left(mol\right)\)

\(\Rightarrow m_{C_2H_6O}=0,072.46=3,312\left(g\right)\)

\(\Rightarrow V_{C_2H_6O}=\dfrac{3,312}{0,8}=4,14\left(ml\right)\)

Độ rượu = \(\dfrac{4,14}{4,5}.100=92^o\)

Khi tính độ rượu thì mình cần phải ghi thêm % vô nữa hay sao bạn

$a\big)$

$n_{C_2H_5OH}=\frac{18,4}{46}=0,4(mol)$

$C_2H_5OH+3O_2\xrightarrow{t^o}2CO_2+3H_2O$

Theo PT: $n_{O_2}=3n_{C_2H_5OH}=1,2(mol)$

$\to V_{kk}=1,2.5.22,4=134,4(l)$

$b\big)$

$CO_2+Ca(OH)_2\to CaCO_3\downarrow+H_2O$

Theo PT: $n_{CaCO_3}=n_{CO_2}=2n_{C_2H_5OH}=0,8(mol)$

$\to m_{\downarrow}=m_{CaCO_3}=0,8.100=80(g)$