Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, nBr2 = 8/160 = 0,05 (mol)
PTHH: C2H4 + Br2 -> C2H4Br2
Mol: 0,05 <--- 0,05 <--- 0,05
Vhh khí = 2,8/22,4 = 0,125 (mol)
%VC2H4 = 0,05/0,125 = 40%
%CH4 = 100% - 40% = 60%
b, nCH4 = 0,125 - 0,05 = 0,075 (mol)
PTHH: C2H4 + 3O2 -> (t°) 2CO2 + 2H2O
Mol: 0,05 ---> 0,15
CH4 + 2O2 -> (t°) CO2 + 2H2O
Mol: 0,075 ---> 0,15
Vkk = (0,15 + 0,15) . 5 . 22,4 = 33,6 (l)
\(a) C_2H_4 + Br_2 \to C_2H_4Br_2\\ n_{C_2H_4} = n_{Br_2} = \dfrac{48}{160} = 0,3(mol)\\ \%V_{C_2H_4} = \dfrac{0,3.22,4}{8,96}.100\% = 75\%\\ \%V_{CH_4} = 100\% -75\% = 25\%\\ b)\)
Khí còn lại : CH4
\(CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + H_2O\\ n_{CO_2} = n_{CH_4} = \dfrac{8,96.25\%}{22,4} = 0,1(mol)\\ m_{CO_2} = 0,1.44 = 4,4(gam)\)
PTHH: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
a) Ta có: \(n_{Br_2}=\dfrac{4}{160}=0,025\left(mol\right)=n_{C_2H_4}\)
\(\Rightarrow\%V_{C_2H_4}=\dfrac{0,025}{\dfrac{5,6}{22,4}}\cdot100\%=10\%\) \(\Rightarrow\%V_{CH_4}=90\%\)
b) Theo PTHH: \(n_{C_2H_4Br_2}=n_{Br_2}=0,025mol\)
\(\Rightarrow m_{C_2H_4Br_2}=0,025\cdot188=4,7\left(g\right)\)
c) Ta có: \(n_{CH_4}=\dfrac{1}{22,4}=\dfrac{5}{112}\left(mol\right)=n_{O_2}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{CH_4}=\dfrac{5}{112}\cdot16\approx0,71\left(g\right)\\m_{O_2}=\dfrac{5}{112}\cdot32\approx1,43\left(g\right)\end{matrix}\right.\)
Vậy 1 lít Metan nhẹ hơn 1 lít Oxi
nhh khí = 3,36/22,4 = 0,15 (mol)
mBr2 = 200 . 10% = 20 (g)
nBr2 = 20/160 = 0,125 (mol)
PTHH: C2H2 + 2Br2 -> C2H2Br4
Mol: 0,0625 <--- 0,125 ---> 0,0625
%VC2H2 = 0,0625/0,15 = 41,66%
%VCH4 = 100% - 41,66% = 58,34%
nCH4 = 0,15 - 0,0625 = 0,0875 (mol)
PTHH:
2C2H2 + 5O2 -> (t°) 4CO2 + 2H2O
0,0625 ---> 0,15625
CH4 + 2O2 -> (t°) CO2 + 2H2O
0,0875 ---> 0,175
Vkk = 22,4 . (0,175 + 0,15625) . 5 = 37,1 (l)
a, \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
\(n_{Br_2}=\dfrac{8}{160}=0,05\left(mol\right)\)
\(n_{C_2H_2}=\dfrac{1}{2}n_{Br_2}=0,025\left(mol\right)\Rightarrow m_{C_2H_2}=0,025.26=0,65\left(g\right)\)
b, \(\left\{{}\begin{matrix}\%V_{C_2H_2}=\dfrac{0,025.22,4}{33,6}.100\%\approx1,67\%\\\%V_{CH_4}\approx98,33\%\end{matrix}\right.\)
nBr2 = 32/160 = 0,2 (mol)
PTHH: C2H4 + Br2 -> C2H4Br2
Mol: 0,2 <--- 0,2
nhh khí = 44,8/22,4 = 2 (mol)
%VC2H4 = 0,2/2 = 10%
%VCH4 = 100% - 10% = 90%
a.\(m_{Br_2}=8g\)
\(n_{Br_2}=\dfrac{8}{160}=0,05mol\)
\(n_{hh}=\dfrac{4,48}{22,4}=0,2mol\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
0,025 0,05 ( mol )
\(\%V_{C_2H_2}=\dfrac{0,025}{0,2}.100=12,5\%\)
\(\%V_{CH_4}=100\%-12,5\%=87,5\%\)
b.
\(m_{C_2H_2}=0,025.26=0,65g\)
\(m_{CH_4}=\left(0,2-0,025\right).16=2,8g\)
c.
\(2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\)
0,025 0,0625 ( mol )
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
0,175 0,35 ( mol )
\(V_{kk}=V_{O_2}.5=\left(0,35+0,0625\right).22,4.5=46,2l\)