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Theo gt ta có: $n_{hh}=0,08(mol);n_{Br_2}=0,08(mol)$
$C_2H_2+2Br_2\rightarrow C_2H_2Br_4$
Suy ra $n_{C_2H_2}=0,04(mol)=n_{CH_4}$
a, $\Rightarrow \%V_{C_2H_2}=\%V_{C_2H_4}=50\%$
b, $CH_4+2O_2\rightarrow CO_2+2H_2O$
$2C_2H_2+5O_2\rightarrow 4CO_2+2H_2O$
Ta có: $n_{O_2}=0,04.2+0,04.5=0,28(mol)\Rightarrow m_{O_2}=8,96(g)$
\(a)C_2H_2 +2Br_2 \to C_2H_2Br_2\\ n_{C_2H_2} = \dfrac{1}{2}n_{Br_2} = \dfrac{0,4.0,2}{2} = 0,04(mol)\\ \Rightarrow V_{C_2H_2} = 0,04.22,4 = 0,896(lít)\\ \%V_{C_2H_2} =\dfrac{0,896}{1,792}.100\% = 50\%\\ \Rightarrow \%V_{CH_4} = 100\% -50\% = 50\%\\ b)\\V_{CH_4} = V_{C_2H_2} = 0,896(lít)\\ CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ C_2H_2 + \dfrac{5}{2}O_2 \xrightarrow{t^o} 2CO_2 + H_2O\\ \)
\(V_{O_2} = 2V_{CH_4} + \dfrac{5}{2}V_{C_2H_2} = 4,032(lít)\\ \Rightarrow m_{O_2} = \dfrac{4,032}{22,4}.32 = 5,76(gam)\)
\(a,Gọi\left\{{}\begin{matrix}n_{CH_4}=a\left(mol\right)\\n_{C_2H_4}=b\left(mol\right)\\n_{C_2H_2}=c\left(mol\right)\end{matrix}\right.\\ n_{hhkhí}=0,4\left(mol\right)\\ n_{CO_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\\ n_{Br_2}=\dfrac{64}{160}=0,4\left(mol\right)\\ PTHH:C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ Mol:a\rightarrow a\\ C_2H_2+2Br_2\rightarrow C_2H_2Br_4\\ Mol:b\rightarrow2b\\ CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\\ Mol:a\rightarrow2a\rightarrow a\)
\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\\ Mol:b\rightarrow3b\rightarrow2b\\ 2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\\ Mol:c\rightarrow2,5c\rightarrow2c\\ Hệ.pt\left\{{}\begin{matrix}a+b+c=0,4\\b+2c=0,4\\a+2b+2c=0,7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,2\left(mol\right)\\c=0,1\left(mol\right)\end{matrix}\right.\)
\(\%V_{CH_4}=\%V_{C_2H_2}=\dfrac{0,1}{0,4}=25\%\\ \%V_{C_2H_4}=\dfrac{0,2}{0,4}=50\%\)
\(m_{CH_4}=0,1.16=1,6\left(g\right)\\ m_{C_2H_4}=28.0,2=5,6\left(g\right)\\ m_{C_2H_2}=0,1.26=2,6\left(g\right)\\ \%m_{CH_4}=\dfrac{1,6}{1,6+5,6+2,6}=16,32\%\\ \%m_{C_2H_4}=\dfrac{5,6}{1,6+5,6+2,6}=57,14\%\\ \%m_{C_2H_2}=100\%-16,32\%-57,14\%=26,54\%\)
\(b,PTHH:C_2H_5OH\rightarrow C_2H_4+H_2O\\ Mol:0,2\leftarrow0,2\\ m_{C_2H_5OH}=0,2.46=9,2\left(g\right)\)
Dài quá!!!
a)
\(n_{Br_2}=\dfrac{8}{160}=0,05\left(mol\right)\)
PTHH: C2H4 + Br2 --> C2H4Br2
0,05<-0,05
=> \(n_{CH_4}=\dfrac{3,36}{22,4}-0,05=0,1\left(mol\right)\)
\(\%m_{CH_4}=\dfrac{0,1.16}{0,1.16+0,05.28}.100\%=53,33\%\)
\(\%m_{C_2H_4}=\dfrac{0,05.28}{0,1.16+0,05.28}.100\%=46,67\%\)
b)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
0,1-->0,2
C2H4 + 3O2 --to--> 2CO2 + 2H2O
0,05--->0,15
=> \(V_{O_2}=\left(0,2+0,15\right).22,4=7,84\left(l\right)\)
a)
Khí còn lại là CH4
\(n_{CH_4}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)
=> \(n_{C_2H_4}=\dfrac{1,16-0,02.16}{28}=0,03\left(mol\right)\)
\(\%V_{CH_4}=\dfrac{0,02}{0,02+0,03}.100\%=40\%\)
\(\%V_{C_2H_4}=\dfrac{0,03}{0,02+0,03}.100\%=60\%\)
b)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
0,02-------------->0,02
C2H4 + 3O2 --to--> 2CO2 + 2H2O
0,03------------->0,06
=> nCO2 = 0,02 + 0,06 = 0,08 (mol)
PTHH: Ca(OH)2 + CO2 --> CaCO3 + H2O
0,08----->0,08
=> mCaCO3 = 0,08.100 = 8 (g)
\(a,n_{hhkhí\left(C_2H_4,C_2H_2\right)}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{Br_2}=\dfrac{80}{160}=0,5\left(mol\right)\\ Gọi\left\{{}\begin{matrix}n_{C_2H_4}=a\left(mol\right)\\n_{C_2H_2}=b\left(mol\right)\end{matrix}\right.\\ PTHH:C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ Mol:a\rightarrow a\\ C_2H_2+2Br_2\rightarrow C_2H_2Br_4\\ Mol:b\rightarrow2b\\ Hệ.pt\left\{{}\begin{matrix}a+b=0,3\\a+2b=0,5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\\ \%V_{C_2H_4}=\dfrac{0,1}{0,3}=33,33\%\\ \%V_{C_2H_2}=100\%-33,335=66,67\%\)
\(b,PTHH:\\ C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\\ Mol:0,1\rightarrow0,3\rightarrow0,2\\ 2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\\ Mol:0,2\rightarrow0,25\rightarrow0,4\\ n_{CO_2}=0,2+0,4=0,6\left(mol\right)\\ PTHH:Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\\ Mol:0,6\rightarrow0,6\rightarrow0,6\\ m_{CaCO_3}=0,6.100=60\left(g\right)\)
- Phần 1 tác dụng với Br2: nBr2 = 16:160 = 0,1 mol
C2H2 + 2Br2 → C2H2Br4
0,05 ← 0,1 (mol)
- Đặt số mol khí mỗi phần như sau:
+ Số mol hỗn hợp khí X là: nX = 11,2:22,4 = 0,5 mol
Ta có: n khí P1 + n khí P2 = nX => 0,05 + x + 0,05k + kx = 0,5 <=> (x + 0,05)k = 0,45 - x
=> 
+ Đốt cháy phần 2:
C2H2 + 2,5O2 → t ∘ 2CO2 + H2O
0,05k → 0,1k→ 0,05k (mol)
CH4 + 2O2 → t ∘ CO2 + 2H2O
kx → kx → 2kx (mol)
Sản phẩm cháy gồm
dẫn qua dung dịch Ba(OH)2 dư:
CO2 + Ba(OH)2 → BaCO3↓ + H2O
k(x+0,1) → k(x+0,1) (mol)
Khối lượng dung dịch giảm: m dd giảm = mBaCO3 – mCO2 – mH2O
=> 197k(x+0,1) – 44k(x+0,1) – 18k(2x+0,05) = 69,525
=> 153k(x+0,1) – 18k(2x+0,05) = 69,525
=> k(117x+14,4) = 69,525
CaC2 + 2H2O → Ca(OH)2 + C2H2
0,2 ← 0,2 (mol)
Al4C3 + 12H2O → 4Al(OH)3 + 3CH4
0,1 ← 0,1 (mol)
Giá trị của m là: m = mCaC2 + mAl4C3 = 0,2.64 + 0,1.144 = 27,2 gam
Phần trăm thể tích các khí trong X là:
tác dụng với AgNO3 trong NH3 dư:
C2H2 + 2AgNO3 + 2NH3 → Ag2C2↓ + 2NH4NO3
0,15 → 0,15 (mol)
Khối lượng kết tủa thu được là: mAg2C2 = 0,15.240 = 36 gam
a) \(n_{Br_2\left(p\text{ư}\right)}=\dfrac{6,4}{160}=0,04\left(mol\right);n_{hh}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
PTHH: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
0,04<--0,04
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,04}{0,6}.100\%=6,67\%\\\%V_{CH_4}=100\%-6,67\%=93,33\%\end{matrix}\right.\)
b) \(n_{CH_4}=0,6-0,04=0,56\left(mol\right)\)
PTHH: \(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\)
0,56----------->0,56
\(C_2H_4+3O_2\xrightarrow[]{t^o}2CO_2+2H_2O\)
0,04----------->0,08
\(\Rightarrow V_{CO_2}=\left(0,08+0,56\right).22,4=14,336\left(l\right)\)
CH4+2O2-to>CO2+2H2O
x-----------------------------2x
C2H2+\(\dfrac{5}{2}\)O2-to>2CO2+H2O
y-----------------------------------y
=>\(\left\{{}\begin{matrix}x+y=\dfrac{3,36}{22,4}\\2x+y=\dfrac{4,5}{18}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=0,1\\y=0,05\end{matrix}\right.\)
=>%VCH4=\(\dfrac{0,1.22,4}{3,36}\).100=66,67%
=>%VC2H2=100-66,67%=33,33%
b)
C2H2+2Br2->C2H2Br4
0,05-----0,1 mol
=>m Br2=0,1.160=16g