Chương V. Dẫn xuất của Hiđrocacbon. Polime

a)`C_2H_2 +5/2O_2->2CO_2+H_2O`

0,25------->0,625mol

`n_(C_2H_2)= (5,6)/(22,4)=0,25 mol`

`-> V_(O_2)=0,625.22,4=14l`

b) `->V_(kk)=14.5=70l`

c) `C_2H_2+HBr->C_2H_3Br`(to,xt)

     0,1<------------------------0,1 mol

`C_2H_3Br+HBr-> C_2H_4Br_2`(to,xt)

0,1<--------------------------0,1 mol

`n_(C_2H_4Br_2)=(18,8)/188 =0,1 mol`

`H=80%`

`->C_2H_2=0,1.26.80%= 20,8g`

#Banhoctot.

\(C_{n+1}H_{2n+2}O_2+\dfrac{3n+1}{2}O_2\rightarrow\left(t^o\right)\left(n+1\right)CO_2+\left(n+1\right)H_2O\\ n_{CO_2}=0,3\left(mol\right)\Rightarrow n_{axit}=\dfrac{0,3}{n+1}=\dfrac{9}{14n+46}\\ \Leftrightarrow0,3.\left(14n+46\right)=9\left(n+1\right)\\ \Leftrightarrow4,2n+13,8=9n+9\\ \Leftrightarrow4,8n=4,8\\ \Leftrightarrow n=1\\ Vậy:CTPT.axit:C_2H_6O_2\)

\(Đặt:m_{CuSO_4\left(thêm\right)}=a\left(g\right)\\ V_{ddCuSO_4\left(sau\right)}=\dfrac{a+160}{1,25}\\ C_{MddCuSO_4\left(sau\right)}=1\left(M\right)\\ \Leftrightarrow\dfrac{\dfrac{a+160.5\%}{160}}{\dfrac{a+160}{1,25.1000}}=1\\ \Leftrightarrow a=14,312\left(g\right)\\ \Rightarrow m_{CuSO_4\left(thêm\right)}=14,312\left(g\right)\)

\(a,C_2H_5OH+O_2\left(men.giấm\right)\rightarrow CH_3COOH+H_2O\\ V_{C_2H_5OH\left(ng.chất\right)}=\dfrac{2,875}{10}=0,2875\left(l\right)=287,5\left(ml\right)\\ m_{C_2H_5OH}=287,5.0,8=230\left(g\right)\\ n_{C_2H_5OH}=\dfrac{230}{46}=5\left(mol\right)\\ n_{CH_3COOH\left(LT\right)}=n_{C_2H_5OH}=5\left(mol\right)\\ n_{CH_3COOH\left(TT\right)}=5.80\%=4\left(mol\right)\\ m_{CH_3COOH\left(TT\right)}=4.60=240\left(g\right)\\ b,m_{dd.giấm}=\dfrac{240.100}{5}=4800\left(gam\right)\)

Em nhập lại nội dung câu hỏi nha

Bạn xem lại đề nhé, 2 chất tham gia pư là axit và bazo mà nhỉ?

a, \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

Ta có: \(n_{Br_2}=\dfrac{16}{160}=0,1\left(mol\right)\)

Theo PT: \(n_{C_2H_4Br_2}=n_{Br_2}=0,1\left(mol\right)\Rightarrow m_{C_2H_4Br_2}=0,1.188=18,8\left(g\right)\)

b, Theo PT: \(n_{C_2H_4}=n_{Br_2}=0,1\left(mol\right)\Rightarrow n_{ankan}=\dfrac{6,72}{22,4}-0,1=0,2\left(mol\right)\)

Gọi CTPT của ankan là C_nH_2n+2.

PT: \(C_nH_{2n+2}+\dfrac{3n+1}{2}O_2\underrightarrow{t^o}nCO_2+\left(n+1\right)H_2O\)

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

Theo PT: \(n_{H_2O}=\left(n+1\right)n_{C_nH_{2n+2}}+2n_{C_2H_4}=\left(n+1\right).0,2+2.0,1=\dfrac{14,4}{18}\)

\(\Rightarrow n=2\)

Vậy: CTPT cần tìm là C₂H₆

Câu 1:

\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(n_{NaOH}=0,2.2=0,4\left(mol\right)\)

\(\Rightarrow\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,4}{0,2}=2\)

Vậy: Pư tạo muối trung hòa, NaOH hết.

Đáp án: B

Câu 2: Đáp án: C

PT: \(CuO+CO\underrightarrow{t^o}Cu+CO_2\)

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

Ta có: \(n_{C_2H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PT: \(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)

Theo PT: \(n_{O_2}=\dfrac{5}{2}n_{C_2H_2}=1\left(mol\right)\Rightarrow V_{O_2}=1.22,4=22,4\left(l\right)\)

Đáp án: A