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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)
\(n_{hh}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)
\(n_{Br_2}=\dfrac{16}{160}=0.1\left(mol\right)\)
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(0.1.........0.1\)
\(n_{CH_4}=0.2-0.1=0.1\left(mol\right)\)
\(\%V_{C_2H_4}=\%V_{CH_4}=\dfrac{0.1}{0.2}\cdot100\%=50\%\)
Em cung cấp thông tin đề câu b lại nhé !
a) C2H4 + Br2 --> C2H4Br2
b) \(n_{Br_2}=\dfrac{24}{160}=0,15\left(mol\right)\)
PTHH: C2H4 + Br2 --> C2H4Br2
0,15<--0,15----->0,15
=> \(\%V_{C_2H_4}=\dfrac{0,15.22,4}{7,84}.100\%=42,857\%\)
=> \(\%V_{CH_4}=\dfrac{7,84-0,15.22,4}{7,84}.100\%=57,143\%\)
c) mC2H4Br2 = 0,15.188 = 28,2 (g)
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) mtăng = mC2H4
=> \(n_{C_2H_4}=\dfrac{5,6}{28}=0,2\left(mol\right)\)
=> \(\%V_{C_2H_4}=\dfrac{0,2.22,4}{13,44}.100\%=33,33\%\)
\(\%V_{CH_4}=100\%-33,33\%=66,67\%\)
b) \(n_{CH_4}=\dfrac{13,44.66,67\%}{22,4}=0,4\left(mol\right)\)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
0,4--------------->0,4
C2H4 + 3O2 --to--> 2CO2 + 2H2O
0,2----------------->0,4
Ca(OH)2 + CO2 --> CaCO3 + H2O
0,8----->0,8
=> mCaCO3 = 0,8.100 = 80 (g)
a.\(m_{tăng}=m_{C_2H_4}=5,6g\)
\(n_{hh}=\dfrac{13,44}{22,4}=0,6mol\)
\(n_{C_2H_4}=\dfrac{5,6}{28}=0,2mol\)
\(\%V_{C_2H_4}=\dfrac{0,2}{0,6}.100=33,33\%\)
\(\%V_{CH_4}=100\%-33,33\%=66,67\%\)
b.\(C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\)
0,2 0,4 ( mol )
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
0,4 0,4 ( mol )
\(n_{CO_2}=0,4+0,4=0,8mol\)
\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)
0,8 0,8 ( mol )
\(m_{CaCO_3}=0,8.100=80g\)
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
\(a)\\ C_2H_4 + Br_2 \to C_2H_4Br_2\\ n_{C_2H_4} = n_{Br_2} = \dfrac{80.10\%}{160} = 0,05(mol)\\ \Rightarrow \%m_{C_2H_4} = \dfrac{0,05.28}{2}.100\% = 70\%\\ \%m_{CH_4} = 100\% - 70\% = 30\%\)
\(b)\) Sản phẩm : Đibrom etan
\(n_{C_2H_4Br_2} = n_{Br_2} = 0,05(mol)\\ \Rightarrow m_{C_2H_4Br_2} = 0,05.188 = 9,4\ gam\)