Dẫn 4,48 lít hỗn hợp khí metan CH4 và axetilen C2H2 (ở đktc) đi qua dung dịch brom dư, thấy có 16 gam brom phản ứng.a.

Câu hỏi:

Dẫn 4,48 lít hỗn hợp khí metan CH₄ và axetilen C₂H₂ (ở đktc) đi qua dung dịch brom dư, thấy có 16 gam brom phản ứng.

a. Viết phương trình phản ứng xảy ra.

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b. Tính thành phần phần trăm theo thể tích các khí có trong hỗn hợp.

c. Tính thể tích khí oxi ở đktc cần dùng để đốt cháy hoàn toàn hỗn hợp trên.

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaqGUb % WaaSbaaSqaaiaabIgacaqGObaabeaakiaab2dacaqGGaGaaeiiaiaa % b6gadaWgaaWcbaGaae4qaiaabIeadaWgaaadbaGaaeinaaqabaaale % qaaOGaaeiiaiaabUcacaqGGaGaaeOBamaaBaaaleaacaqGdbWaaSba % aWqaaiaabkdaaeqaaSGaaeisamaaBaaameaacaqGYaaabeaaaSqaba % GccaqGGaGaaeypaiaabccadaWcaaqaaiaabsdacaqGSaGaaeinaiaa % bIdaaeaacaqGYaGaaeOmaiaabYcacaqG0aaaaiaabccacaqG9aGaae % iiaiaabcdacaqGSaGaaeOmaiaabccacaqGOaGaaeyBaiaab+gacaqG % SbGaaeykaaqaaiaab6gadaWgaaWcbaGaaeOqaiaabkhadaWgaaadba % GaaeOmaaqabaaaleqaaOGaaeypaiaabccadaWcaaqaaiaabgdacaqG % 2aaabaGaaeymaiaabAdacaqGWaaaaiaabccacaqG9aGaaeiiaiaabc % dacaqGSaGaaeymaiaabccacaqGOaGaaeyBaiaab+gacaqGSbGaaeyk % aaaaaa!6785! \begin{array}{l} {{\rm{n}}_{{\rm{hh}}}}{\rm{ = }}{{\rm{n}}_{{\rm{C}}{{\rm{H}}_{\rm{4}}}}}{\rm{ + }}{{\rm{n}}_{{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}}}{\rm{ = }}\frac{{{\rm{4,48}}}}{{{\rm{22,4}}}}{\rm{ = 0,2 (mol)}}\\ {{\rm{n}}_{{\rm{B}}{{\rm{r}}_{\rm{2}}}}}{\rm{ = }}\frac{{{\rm{16}}}}{{{\rm{160}}}}{\rm{ = 0,1 (mol)}} \end{array}\)

a, PTHH :   C₂H₂   +   2Br₂  →   C₂H₂Br₄

Theo pthh ta có : 

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaqGUb % WaaSbaaSqaaiaaboeadaWgaaadbaGaaeOmaaqabaWccaqGibWaaSba % aWqaaiaabkdaaeqaaaWcbeaakiaab2dacaqGGaWaaSaaaeaacaqGXa % aabaGaaeOmaaaacaqGUbWaaSbaaSqaaiaabkeacaqGYbWaaSbaaWqa % aiaabkdaaeqaaaWcbeaakiaabccacaqG9aGaaeiiamaalaaabaGaae % imaiaabYcacaqGXaaabaGaaeOmaaaacaqGGaGaaeypaiaabccacaqG % WaGaaeilaiaabcdacaqG1aGaaeiiaiaabIcacaqGTbGaae4BaiaabY % gacaqGPaaabaGaaeOBamaaBaaaleaacaqGdbGaaeisamaaBaaameaa % caqG0aaabeaaaSqabaGccaqG9aGaaeiiaiaabccacaqGUbWaaSbaaS % qaaiaabIgacaqGObaabeaakiaabccacaqGTaGaaeiiaiaab6gadaWg % aaWcbaGaae4qamaaBaaameaacaqGYaaabeaaliaabIeadaWgaaadba % GaaeOmaaqabaaaleqaaOGaaeiiaiaab2dacaqGGaGaaeimaiaabYca % caqGYaGaaeiiaiaab2cacaqGGaGaaeimaiaabYcacaqGWaGaaeynai % abg2da9iaaicdacaGGSaGaaGymaiaaiwdacaqGGaGaaeikaiaab2ga % caqGVbGaaeiBaiaabMcaaeaacaqGwbWaaSbaaSqaaiaadoeadaWgaa % adbaGaaGOmaaqabaWccaWGibWaaSbaaWqaaiaaikdaaeqaaaWcbeaa % kiaab2dacaqGGaGaaeOBamaaBaaaleaadaWgaaadbaGaam4qamaaBa % aabaGaaGOmaaqabaGaamisamaaBaaabaGaaGOmaaqabaaabeaaaSqa % baGccaqGGaGaaeOlaiaabkdacaqGYaGaaeilaiaabsdacaqGGaGaae % ypaiaabccacaqGWaGaaeilaiaabcdacaqG1aGaaeiiaiaab6cacaqG % GaGaaeOmaiaabkdacaqGSaGaaeinaiaabccacaqG9aGaaeiiaiaabg % dacaqGSaGaaeymaiaabkdacaqGGaGaaeikaiaabYgacaqGPaaabaGa % aeyjaiaabAfadaWgaaWcbaGaae4qamaaBaaameaacaqGYaaabeaali % aabIeadaWgaaadbaGaaeOmaaqabaaaleqaaOGaaeypaiaabccadaWc % aaqaaiaabgdacaqGSaGaaeymaiaabkdaaeaacaqG0aGaaeilaiaabs % dacaqG4aaaaiaabccacaqGUaGaaeiiaiaabgdacaqGWaGaaeimaiaa % bwcacaqGGaGaaeypaiaabccacaqGYaGaaeynaiaabwcaaeaacaqGLa % GaaeOvamaaBaaaleaacaqGdbGaaeisamaaBaaameaacaaI0aaabeaa % aSqabaGccqGH9aqpcaaIXaGaaGimaiaaicdacaGGLaGaeyOeI0IaaG % OmaiaaiwdacaGGLaGaeyypa0JaaG4naiaaiwdacaGGLaaaaaa!B3FE! \begin{array}{l} {{\rm{n}}_{{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}{{\rm{n}}_{{\rm{B}}{{\rm{r}}_{\rm{2}}}}}{\rm{ = }}\frac{{{\rm{0,1}}}}{{\rm{2}}}{\rm{ = 0,05 (mol)}}\\ {{\rm{n}}_{{\rm{C}}{{\rm{H}}_{\rm{4}}}}}{\rm{ = }}{{\rm{n}}_{{\rm{hh}}}}{\rm{ – }}{{\rm{n}}_{{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}}}{\rm{ = 0,2 – 0,05}} = 0,15{\rm{ (mol)}}\\ {{\rm{V}}_{{C_2}{H_2}}}{\rm{ = }}{{\rm{n}}_{_{{C_2}{H_2}}}}{\rm{ }}{\rm{.22,4 = 0,05 }}{\rm{. 22,4 = 1,12 (l)}}\\ {\rm{\% }}{{\rm{V}}_{{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{2}}}}}{\rm{ = }}\frac{{{\rm{1,12}}}}{{{\rm{4,48}}}}{\rm{ }}{\rm{. 100\% = 25\% }}\\ {\rm{\% }}{{\rm{V}}_{{\rm{C}}{{\rm{H}}_4}}} = 100\% – 25\% = 75\% \end{array}\)

b, PTHH phản ứng đốt cháy hỗn hợp khí trên :

CH₄  +  2O₂ → CO₂  +  2H₂O   (1)

 0,15      0,3                               

2C₂H₂  +  5O₂ → 4CO₂  +  2H₂O   (2)

0,05      0,125             

Từ pthh (1) và (2) ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeOBamaaBa % aaleaacaqGpbWaaSbaaWqaaiaabkdaaeqaaaWcbeaakiaab2dacaqG % GaGaaeimaiaabYcacaqGZaGaaeiiaiaabUcacaqGGaGaaeimaiaabY % cacaqGXaGaaeOmaiaabwdacaqGGaGaaeypaiaabccacaqGWaGaaeil % aiaabsdacaqGYaGaaeynaiaabccacaqGOaGaaeyBaiaab+gacaqGSb % Gaaeykaaaa!4C22! {{\rm{n}}_{{{\rm{O}}_{\rm{2}}}}}{\rm{ = 0,3 + 0,125 = 0,425 (mol)}}\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaeOvamaaBa % aaleaacaWGpbWaaSbaaWqaaiaaikdaaeqaaaWcbeaakiaab2dacaqG % GaGaaeOBamaaBaaaleaacaqGpbWaaSbaaWqaaiaabkdaaeqaaaWcbe % aakiaabccacaqGUaGaaeOmaiaabkdacaqGSaGaaeinaiaabccacaqG % 9aGaaeiiaiaabcdacaqGSaGaaeinaiaabkdacaqG1aGaaeiiaiaab6 % cacaqGGaGaaeOmaiaabkdacaqGSaGaaeinaiaabccacaqG9aGaaeii % aiaabMdacaqGSaGaaeynaiaabkdacaqGGaGaaeikaiaabYgacaqGPa % aaaa!5351! {{\rm{V}}_{{O_2}}}{\rm{ = }}{{\rm{n}}_{{{\rm{O}}_{\rm{2}}}}}{\rm{ }}{\rm{.22,4 = 0,425 }}{\rm{. 22,4 = 9,52 (l)}}\)