\(a)n_{Zn}=\dfrac{13}{65}=0,2mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,2       0,4              0,2         0,2

\(V_{H_2,đkc}=0,2.24,79=4,958l\\ b)m_{ZnCl_2}=0,2.136=27,2g\\ c)V_{ddHCl}=\dfrac{0,4}{2}=0,2l\)

 Ta có Cùng điều kiện -> Quy số lít về số mol.n(hh ban đầu) = 20 mol; n(hh sau) = 16 lít

=> H2 phản ứng mất 4 lít => C2H2 có 2 lít và CH4 có 8 lít

a.\(n_{hh}=\dfrac{V_{hh}}{22,4}=\dfrac{0,896}{22,4}=0,04mol\)

\(n_{Br_2}=\dfrac{m_{Br_2}}{M_{Br_2}}=\dfrac{8}{160}=0,05mol\)

Gọi \(n_{C_2H_2}\) là x => \(V_{C_2H_2}=22,4x\)

       \(n_{C_2H_4}\) là y => \(V_{C_2H_4}=22,4y\)

\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

  x           2x                 x         ( mol )

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

   y         y                y            ( mol )

Ta có hpt:

\(\left\{{}\begin{matrix}22,4x+22,4y=0,896\\2x+y=0,05\end{matrix}\right.\)   \(\Leftrightarrow\left\{{}\begin{matrix}x=0,01\\y=0,03\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}V_{C_2H_2}=0,01.22,4=0,224\\V_{C_2H_4}=0,03.22,4=0,672\end{matrix}\right.\)

\(\%V_{C_2H_2}=\dfrac{0,224}{0,896}.100=25\%\)

\(\%V_{C_2H_4}=100-25=75\%\)

b.\(m_{Sp}=0,01.186+0,03.346+0,03.188=16,02g\)

( Em cảm ơn anh Kudo Sinichi rất nhiều ạ!! )

\(n_{hh}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

\(n_{Br_2}=0.1\cdot2=0.2\left(mol\right)\)

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

\(0.2..........0.2\)

\(n_{CH_4}=0.3-0.2=0.1\left(mol\right)\)

Câu b anh nghĩ phải là đốt cháy sau đó dẫn sản phẩm vào Ba(OH)₂ dư nha .

\(CH_4+2O_2\underrightarrow{t^0}CO_2+2H_2O\)

\(0.1.....................0.1\)

\(Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3+H_2O\)

\(.............0.1.......0.1\)

\(m_{BaCO_3}=0.1\cdot197=19.7\left(g\right)\)

\(n_{Ca}=\dfrac{12}{40}=0,3\left(mol\right)\)

PT: \(Ca+H_2SO_4\rightarrow CaSO_4+H_2\)

Theo PT: \(n_{H_2SO_4}=n_{CaSO_4}=n_{H_2}=n_{Ca}=0,3\left(mol\right)\)

a, \(V_{H_2}=0,3.24,79=7,437\left(l\right)\)

b, \(m_{CaSO_4}=0,3.136=40,8\left(g\right)\)

c, \(V_{H_2SO_4}=\dfrac{0,3}{1,6}=0,1875\left(l\right)\)

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

b, \(n_{C_2H_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

Theo PT: \(n_{C_2H_4Br_2}=n_{C_2H_4}=0,4\left(mol\right)\Rightarrow m_{C_2H_4Br_2}=0,4.188=75,2\left(g\right)\)

c, \(n_{Br_2}=n_{C_2H_4}=0,4\left(mol\right)\Rightarrow C_{M_{Br_2}}=\dfrac{0,4}{0,2}=2\left(M\right)\)

PTHH: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

Ta có: \(\left\{{}\begin{matrix}n_{hhkhí}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\n_{C_2H_4Br_2}=\dfrac{18,8}{188}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{C_2H_4}=n_{CH_4}=n_{Br_2}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}V_{CH_4}=V_{C_2H_4}=0,1\cdot22,4=2,24\left(l\right)\\m_{Br_2}=0,1\cdot160=16\left(g\right)\end{matrix}\right.\)

PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

a, Ta có: \(n_{hhk}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

 \(n_{C_2H_4Br_2}=\dfrac{18,8}{188}=0,1\left(mol\right)\)

Theo PT: \(n_{C_2H_4}=n_{C_2H_4Br_2}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,1}{0,2}.100\%=50\%\\\%V_{CH_4}=50\%\end{matrix}\right.\)

b, Theo PT: \(n_{Br_2}=n_{C_2H_4Br_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{Br_2}=0,1.160=16\left(g\right)\)

Bạn tham khảo nhé!

mb` tăng = mC2H2= 5.2g=>nC2H2=0.2(mol)

C2H2 + 2Br2 --> C2H2Br4

0.2-------0.4----------0.2

=>VBr2=0.4/2=0.2(l)

n_hhkhí=8.96/22.4=0.4(mol)==> nCH4= 0.4-0.2=0.2(mol)

=>%CH4=0.2x22.4x100/8.96=50%

=>%C2H2=100%-50%=50%

Chỉ có C2H2 pư với Brom

Nên khối lượng bình tăng 5.2g là khối lượng của C2H2:

=> nC2H2= 5.2/26=0.2 mol

PTHH : C2H2 + Br2 ---> C2H2Br2

0.2 0.2

V_Br2= n/CM = 0.2*2 = 0.4 lít

VC2H2= 0.2*22.4= 4.48 lít => %VC2H2= \(\frac{4.48}{8.96}\cdot100\) = 50%

=> %VCH4 = 100% - 50% = 50%

(có trường hợp : C2H2 + Br2 ---> C2H2Br4 .... lên lớp trên sẽ rõ hơn nhé :)

Chúc em học tốt !!

\(m_{tăng}=m_{C_2H_4}=5.6\left(g\right)\)

\(m_{CH_4}=8.8-5.6=3.2\left(g\right)\)

\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

\(\%CH_4=\dfrac{3.2}{8.8}\cdot100\%=36.36\%\)

\(\%C_2H_4=100-36.36=63.64\%\)

\(m_{C_2H_4Br_2}=188\cdot0.2=37.6\left(g\right)\)

\(n_{Br_2} = \dfrac{4,8}{160} = 0,03(mol)\\ C_2H_2 + 2Br_2 \to C_2H_2Br_4\\ n_{C_2H_2}= \dfrac{1}{2}n_{Br_2} = 0,015(mol)\\ \%V_{C_2H_2} = \dfrac{0,015.22,4}{1,12}.100\% = 30\%\\ \%V_{CH_4} = 100\% -30\%=70\%\)