\(a.C_2H_5OH+3O_2-^{t^o}\rightarrow2CO_2+3H_2O\\ n_{C_2H_5OH}=0,3\left(mol\right)\\ n_{CO_2}=2n_{C_2H_5OH}=0,6\left(mol\right)\\ \Rightarrow V_{CO_2}=0,6.22,4=13,44\left(l\right)\\ b.n_{O_2}=3n_{C_2H_5OH}=0,6\left(mol\right)\\ MàV_{O_2}=\dfrac{1}{5}V_{kk}\\ \Rightarrow V_{kk}=V_{O_2}.5=0,6.22,4.5=67,2\left(l\right)\\ c.n_{NaOH}=0,9\left(mol\right)\\ Tacó:\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,9}{0,6}=1,5\\ \Rightarrow Tạora2muốiNaHCO_3vàNa_2CO_3\\ Đặt:n_{NaHCO_3}=x\left(mol\right);n_{Na_2CO_3}=y\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}x+y=0,6\left(BTnguyento\left(C\right)\right)\\x+2y=0,9\left(BTnguyento\left(Na\right)\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,3\\y=0,3\end{matrix}\right.\\ \Rightarrow m_{muối}=0,3.84+0,3.106=57\left(g\right)\)

a)C2H6O+3O2to→2CO2+3H2O

0,3-------------0,9------0,6

nC2H6O=\(\dfrac{13,8}{46}\)=0,3mol

nCO2=2nC2H6O=0,6mol

⇒VCO2=0,6×22,4=13,44l

⇒Vkk=0,9×22,4×5=100,8l

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

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

\(V_{C_2H_5OH}=\dfrac{20.96}{100}=19,2\left(ml\right)\)

=> \(m_{C_2H_5OH}\) = 19,2.0,8 = 15,36 (g)

=> \(n_{C_2H_5OH}=\dfrac{15,36}{46}=\dfrac{192}{575}\left(mol\right)\) 

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

Theo PTHH: \(n_{O_2}=\dfrac{576}{575}\left(mol\right)\Rightarrow V_{O_2\left(đktc\right)}=\dfrac{576}{575}.22,4=\dfrac{12902,4}{575}\left(l\right)\)

a, \(n_{C_2H_6O}=\dfrac{92}{46}=2\left(mol\right)\)

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

Theo PT: \(n_{O_2}=3n_{C_2H_6O}=6\left(mol\right)\Rightarrow V_{O_2}=6.22,4=134,4\left(l\right)\)

b, \(V_{kk}=5V_{O_2}=672\left(l\right)\)

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 = \(\dfrac{28,75}{30}.100=95,83^o\)

Đáp án: D

Thể tích ứng với 16 gam rượu là → V= 16: 0,8 = 20ml

Vậy dung dịch có độ rượu là : 20/250 ×100 = 8⁰

\(n_{CaCO_3}=\dfrac{100,2}{100}=1,002\left(mol\right)\)

PTHH: Ca(OH)₂ + CO₂ ---> CaCO₃ + H₂O

                             1,002 <---- 1,002

            C₂H₅OH + 3O₂ --t^o--> 2CO₂ + 3H₂O

             0,501 <-------------------- 1,002

\(\rightarrow m_{C_2H_5OH}=0,501.46=23,046\left(g\right)\\ \rightarrow V_{C_2H_5OH}=\dfrac{23,046}{0,8}=28,8075\left(ml\right)\)

=> Độ rượu là: \(\dfrac{29,8075}{30}=96,025^o\)