Câu hỏi của ✎﹏トラン⋮ Hannie ッ

Question

Cho 3,584 lít ( đktc ) hỗn hợp A gồm một ankan ( X ), một anken ( Y ), một ankin ( Z ).-Lấy một nửa hh A cho t/d với dd AgNO3 dư trong amoniac thấy thể tích hh giảm 12,5% và thu được 2,4 gam kết tủa.-...

๖ۣۜDũ๖ۣۜN๖ۣۜG · Accepted Answer

a)CTPT của X, Y, Z là CxH2x+2, CyH2y, CzH2z-2$C_zH_{2z-2}+AgNO_3+NH_3\rightarrow C_zH_{2z-3}Ag\downarrow+NH_4NO_3$$C_yH_{2y}+Br_2\rightarrow C_yH_{2y}Br_2$$C_zH_{2z-2}+2Br_2\rightarrow C_zH_{2z-2}Br_4$$C_xH_{2x+2}+\dfrac{3x+1}{2}O_2\underrightarrow{t^o}xCO_2+\left(x+1\right)H_2O$$Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3+H_2O$b)$n_{C_zH_{2z-2}}=\dfrac{1,792.12,5\%}{22,4}=0,01\left(mol\right)$- Nếu Z là C2H2:PTHH: $C_2H_2+2AgNO_3+2NH_3\rightarrow C_2Ag_2+2NH_4NO_3$ 0,01---------------------------->0,01=> mkt = 0,01.240 = 2,4 (g) => Thỏa mãn $n_{Br_2}=\dfrac{13,6}{160}=0,085\left(mol\right)$PTHH: $C_2H_2+2Br_2\rightarrow C_2H_2Br_4$ 0,01---->0,02 $C_yH_{2y}+Br_2\rightarrow C_yH_{2y}Br_2$ 0,065<--0,065=> mtăng = 0,01.26 + 0,065.14y = 2,99=> y = 3=> Y là C3H6$n_X=\dfrac{1,792}{22,4}-0,01-0,065=0,005\left(mol\right)$$n_{BaCO_3}=\dfrac{3,94}{197}=0,02\left(mol\right)$Bảo toàn C: 0,005x = 0,02=> x = 4 => X là C4H10$\left\{{}\begin{matrix}\%V_{C_2H_2}=\dfrac{0,01.22,4}{1,792}.100\%=12,5\%\\%V_{C_3H_6}=\dfrac{0,065.22,4}{1,792}.100\%=81,25\%\\%V_{C_4H_{10}}=\dfrac{0,005.22,4}{1,792}.100\%=6,25\%\end{matrix}\right.$- Nếu Z không phải là C2H2PTHH: $C_zH_{2z-2}+AgNO_3+NH_3\rightarrow C_zH_{2z-3}Ag+NH_4NO_3$ 0,01--------------------------->0,01=> $M_{C_zH_{2z-3}Ag}=\dfrac{2,4}{0,01}=240\left(g/mol\right)$=> LoạiCâu trả lời: a)CTPT của X, Y, Z là CxH2x+2, CyH2y, CzH2z-2$C_zH_{2z-2}+AgNO_3+NH_3\rightarrow C_zH_{2z-3}Ag\downarrow+NH_4NO_3$$C_yH_{2y}+Br_2\rightarrow C_yH_{2y}Br_2$$C_zH_{2z-2}+2Br_2\rightarrow C_zH_{2z-2}Br_4$$C_xH_{2x+2}+\dfrac{3x+1}{2}O_2\underrightarrow{t^o}xCO_2+\left(x+1\right)H_2O$$Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3+H_2O$b)$n_{C_zH_{2z-2}}=\dfrac{1,792.12,5\%}{22,4}=0,01\left(mol\right)$- Nếu Z là C2H2:PTHH: $C_2H_2+2AgNO_3+2NH_3\rightarrow C_2Ag_2+2NH_4NO_3$ 0,01---------------------------->0,01=> mkt = 0,01.240 = 2,4 (g) => Thỏa mãn $n_{Br_2}=\dfrac{13,6}{160}=0,085\left(mol\right)$PTHH: $C_2H_2+2Br_2\rightarrow C_2H_2Br_4$ 0,01---->0,02 $C_yH_{2y}+Br_2\rightarrow C_yH_{2y}Br_2$ 0,065<--0,065=> mtăng = 0,01.26 + 0,065.14y = 2,99=> y = 3=> Y là C3H6$n_X=\dfrac{1,792}{22,4}-0,01-0,065=0,005\left(mol\right)$$n_{BaCO_3}=\dfrac{3,94}{197}=0,02\left(mol\right)$Bảo toàn C: 0,005x = 0,02=> x = 4 => X là C4H10$\left\{{}\begin{matrix}\%V_{C_2H_2}=\dfrac{0,01.22,4}{1,792}.100\%=12,5\%\\%V_{C_3H_6}=\dfrac{0,065.22,4}{1,792}.100\%=81,25\%\\%V_{C_4H_{10}}=\dfrac{0,005.22,4}{1,792}.100\%=6,25\%\end{matrix}\right.$- Nếu Z không phải là C2H2PTHH: $C_zH_{2z-2}+AgNO_3+NH_3\rightarrow C_zH_{2z-3}Ag+NH_4NO_3$ 0,01--------------------------->0,01=> $M_{C_zH_{2z-3}Ag}=\dfrac{2,4}{0,01}=240\left(g/mol\right)$=> Loại

Kudo Shinichi · Answer

a)Gọi CTHH của ankan X, anken Y, ankin Z lần lượt là $C_xH_{2x+2};C_yH_{2y};C_zH_{2z-2}\left(x,y,z\in N\text{*}\right)$$C_zH_{2z-2}+aAgNO_3+aNH_3\rightarrow C_zH_{2z-2-a}Ag_a\downarrow+aNH_4NO_3$$C_yH_{2y}+Br_2\rightarrow C_yH_{2y}Br_2$$C_zH_{2z-2}+2Br_2\rightarrow C_zH_{2z-2}Br_4$$2C_xH_{2x+2}+\left(3x+1\right)O_2\xrightarrow[]{t^o}2xCO_2+\left(2x+2\right)H_2O$$CO_2+Ba\left(OH\right)_2\rightarrow BaCO_3+H_2O$b)- TN1: $\%V_{giảm}=\%V_{C_zH_{2z-2}}=12,5\%$$V_{C_zH_{2z-2}}=12,5\%.3,584=0,448\left(l\right)\ \Rightarrow\dfrac{1}{2}n_{C_zH_{2z-2}}=\dfrac{0,448}{22,4}=0,01\left(mol\right)$PTHH:$C_zH_{2z-2}+aAgNO_3+aNH_3\rightarrow C_zH_{2z-2-a}Ag_a\downarrow+aNH_4NO_3$0,01---------------------------------->0,01=> $M_{C_zH_{2z-2-a}}=\dfrac{2,4}{0,01}=240\left(g/mol\right)$=> $C_2Ag_2$=> z = 2`ankin Z: C_2H_2`- TN2:$n_{Br_2}=\dfrac{13,6}{160}=0,085\left(mol\right);n_{C_2H_2}=0,01\left(mol\right)$$m_{tăng}=m_{C_2H_2}+m_{C_yH_{2y}}\ \Rightarrow m_{C_yH_{2y}}=2,99-0,01.26=2,73\left(g\right)$$C_2H_2+2Br_2\rightarrow C_2H_2Br_4$0,01--->0,02$C_yH_{2y}+Br_2\rightarrow C_yH_{2y}Br_2$0,065<--0,065=> $M_{C_yH_{2y}}=\dfrac{2,73}{0,065}=42\left(g/mol\right)$=> y = 3`anken Y: C_3H_6`$\dfrac{1}{2}n_{hh}=\dfrac{1}{2}.\dfrac{3,584}{22,4}=0,08\left(mol\right);\ \Rightarrow n_{C_xH_{2x+2}}=0,08-0,01-0,065=0,005\left(mol\right)$$n_{CO_2}=n_{BaCO_3}=\dfrac{3,94}{197}=0,02\left(mol\right)$Ta có: $\dfrac{n_{C_xH_{2x+2}}}{n_{CO_2}}=\dfrac{2}{2x}=\dfrac{0,005}{0,02}$=> x = 4=> `ankan X: C_4H_{10}`Câu trả lời: a)Gọi CTHH của ankan X, anken Y, ankin Z lần lượt là $C_xH_{2x+2};C_yH_{2y};C_zH_{2z-2}\left(x,y,z\in N\text{*}\right)$$C_zH_{2z-2}+aAgNO_3+aNH_3\rightarrow C_zH_{2z-2-a}Ag_a\downarrow+aNH_4NO_3$$C_yH_{2y}+Br_2\rightarrow C_yH_{2y}Br_2$$C_zH_{2z-2}+2Br_2\rightarrow C_zH_{2z-2}Br_4$$2C_xH_{2x+2}+\left(3x+1\right)O_2\xrightarrow[]{t^o}2xCO_2+\left(2x+2\right)H_2O$$CO_2+Ba\left(OH\right)_2\rightarrow BaCO_3+H_2O$b)- TN1: $\%V_{giảm}=\%V_{C_zH_{2z-2}}=12,5\%$$V_{C_zH_{2z-2}}=12,5\%.3,584=0,448\left(l\right)\ \Rightarrow\dfrac{1}{2}n_{C_zH_{2z-2}}=\dfrac{0,448}{22,4}=0,01\left(mol\right)$PTHH:$C_zH_{2z-2}+aAgNO_3+aNH_3\rightarrow C_zH_{2z-2-a}Ag_a\downarrow+aNH_4NO_3$0,01---------------------------------->0,01=> $M_{C_zH_{2z-2-a}}=\dfrac{2,4}{0,01}=240\left(g/mol\right)$=> $C_2Ag_2$=> z = 2`ankin Z: C_2H_2`- TN2:$n_{Br_2}=\dfrac{13,6}{160}=0,085\left(mol\right);n_{C_2H_2}=0,01\left(mol\right)$$m_{tăng}=m_{C_2H_2}+m_{C_yH_{2y}}\ \Rightarrow m_{C_yH_{2y}}=2,99-0,01.26=2,73\left(g\right)$$C_2H_2+2Br_2\rightarrow C_2H_2Br_4$0,01--->0,02$C_yH_{2y}+Br_2\rightarrow C_yH_{2y}Br_2$0,065<--0,065=> $M_{C_yH_{2y}}=\dfrac{2,73}{0,065}=42\left(g/mol\right)$=> y = 3`anken Y: C_3H_6`$\dfrac{1}{2}n_{hh}=\dfrac{1}{2}.\dfrac{3,584}{22,4}=0,08\left(mol\right);\ \Rightarrow n_{C_xH_{2x+2}}=0,08-0,01-0,065=0,005\left(mol\right)$$n_{CO_2}=n_{BaCO_3}=\dfrac{3,94}{197}=0,02\left(mol\right)$Ta có: $\dfrac{n_{C_xH_{2x+2}}}{n_{CO_2}}=\dfrac{2}{2x}=\dfrac{0,005}{0,02}$=> x = 4=> `ankan X: C_4H_{10}`

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