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\(n_{hhkhí}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\\ m_{tăng}=m_{C_2H_4}=4,2\left(g\right)\\ n_{C_2H_4}=\dfrac{4,2}{28}=0,15\left(mol\right)\\ n_{CH_4}=0,35-0,15=0,2\left(mol\right)\\ \left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,15}{0,35}=42,85\%\\\%V_{CH_4}=100\%-42,85\%=57,15\%\end{matrix}\right.\)
PTHH:
C2H4 + 3O2 --to--> 2CO2 + 2H2O
0,15 ------------------> 0,3
CH4 + O2 --to--> CO2 + 2H2O
0,2 -----------------> 0,2
Ca(OH)2 + CO2 ---> CaCO3 + H2O
0,5 -------> 0,5
\(m_{CaCO_3}=0,5.100=50\left(g\right)\)
\(n_{\downarrow}=\dfrac{35}{100}=0,35mol\Rightarrow n_C=m_{CaCO_3}=0,35mol\)
\(\left\{{}\begin{matrix}CH_4:x\left(mol\right)\\C_2H_2:y\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+y=\dfrac{4,48}{22,4}=0,2\\BTC:x+2y=0,35\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,15\end{matrix}\right.\)
\(m_{tăng}=m_{Br_2}=2n_{C_2H_2}\cdot160=48g\)
\(\%V_{CH_4}=\dfrac{0,05}{0,05+0,15}\cdot100\%=25\%\)
\(\%V_{C_2H_2}=100\%-25\%=75\%\)
Gọi \(\left\{{}\begin{matrix}n_{CH_4}=x\\n_{C_2H_2}=y\end{matrix}\right.\)
\(n_{hh}=\dfrac{4,48}{22,4}=0,2mol\)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
x x ( mol )
\(2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\)
y 2y ( mol )
\(n_{CaCO_3}=\dfrac{35}{100}=0,35mol\)
\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)
0,35 0,35 ( mol )
Ta có:
\(\left\{{}\begin{matrix}x+y=0,2\\x+2y=0,35\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,05\\y=0,15\end{matrix}\right.\)
\(m_{tăng}=2m_{C_2H_2}=2.0,15.160=48g\)
\(V_{CH_4}=0,05.22,4=1,12l\)
\(V_{C_2H_2}=0,15.22,4=3,36l\)
\(a,Gọi\left\{{}\begin{matrix}n_{CH_4}=a\left(mol\right)\\n_{C_2H_4}=b\left(mol\right)\\n_{C_2H_2}=c\left(mol\right)\end{matrix}\right.\\ n_{hhkhí}=0,4\left(mol\right)\\ n_{CO_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\\ n_{Br_2}=\dfrac{64}{160}=0,4\left(mol\right)\\ PTHH:C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ Mol:a\rightarrow a\\ C_2H_2+2Br_2\rightarrow C_2H_2Br_4\\ Mol:b\rightarrow2b\\ CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\\ Mol:a\rightarrow2a\rightarrow a\)
\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\\ Mol:b\rightarrow3b\rightarrow2b\\ 2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\\ Mol:c\rightarrow2,5c\rightarrow2c\\ Hệ.pt\left\{{}\begin{matrix}a+b+c=0,4\\b+2c=0,4\\a+2b+2c=0,7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,2\left(mol\right)\\c=0,1\left(mol\right)\end{matrix}\right.\)
\(\%V_{CH_4}=\%V_{C_2H_2}=\dfrac{0,1}{0,4}=25\%\\ \%V_{C_2H_4}=\dfrac{0,2}{0,4}=50\%\)
\(m_{CH_4}=0,1.16=1,6\left(g\right)\\ m_{C_2H_4}=28.0,2=5,6\left(g\right)\\ m_{C_2H_2}=0,1.26=2,6\left(g\right)\\ \%m_{CH_4}=\dfrac{1,6}{1,6+5,6+2,6}=16,32\%\\ \%m_{C_2H_4}=\dfrac{5,6}{1,6+5,6+2,6}=57,14\%\\ \%m_{C_2H_2}=100\%-16,32\%-57,14\%=26,54\%\)
\(b,PTHH:C_2H_5OH\rightarrow C_2H_4+H_2O\\ Mol:0,2\leftarrow0,2\\ m_{C_2H_5OH}=0,2.46=9,2\left(g\right)\)
Dài quá!!!
\(a,n_{hhkhí\left(C_2H_4,C_2H_2\right)}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{Br_2}=\dfrac{80}{160}=0,5\left(mol\right)\\ Gọi\left\{{}\begin{matrix}n_{C_2H_4}=a\left(mol\right)\\n_{C_2H_2}=b\left(mol\right)\end{matrix}\right.\\ PTHH:C_2H_4+Br_2\rightarrow C_2H_4Br_2\\ Mol:a\rightarrow a\\ C_2H_2+2Br_2\rightarrow C_2H_2Br_4\\ Mol:b\rightarrow2b\\ Hệ.pt\left\{{}\begin{matrix}a+b=0,3\\a+2b=0,5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\\ \%V_{C_2H_4}=\dfrac{0,1}{0,3}=33,33\%\\ \%V_{C_2H_2}=100\%-33,335=66,67\%\)
\(b,PTHH:\\ C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\\ Mol:0,1\rightarrow0,3\rightarrow0,2\\ 2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\\ Mol:0,2\rightarrow0,25\rightarrow0,4\\ n_{CO_2}=0,2+0,4=0,6\left(mol\right)\\ PTHH:Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\\ Mol:0,6\rightarrow0,6\rightarrow0,6\\ m_{CaCO_3}=0,6.100=60\left(g\right)\)
a)
CH4 + 2O2 --to--> CO2 + 2H2O
C2H4 + 3O2 --to--> 2CO2 + 2H2O
b) Gọi số mol CH4, C2H4 là a, b (mol)
=> \(a+b=\dfrac{6,72}{22,4}=0,3\)
\(n_{CaCO_3}=\dfrac{20}{100}=0,2\left(mol\right)\)
Khí thoát ra khỏi bình là CH4
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
a---------------->a
Ca(OH)2 + CO2 --> CaCO3 + H2O
0,2<------0,2
=> a = 0,2 (mol)
=> \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,2}{0,3}.100\%=66,67\%\\\%V_{C_2H_4}=100\%-66,67\%=33,33\%\end{matrix}\right.\)
c) b = 0,1 (mol)
CH4 + 2O2 --to--> CO2 + 2H2O
0,2--------------->0,2----->0,4
C2H4 + 3O2 --to--> 2CO2 + 2H2O
0,1----------------->0,2---->0,2
Ca(OH)2 + CO2 --> CaCO3 + H2O
0,4------>0,4
=> \(m_{CaCO_3}=0,4.100=40\left(g\right)\)
\(\left\{{}\begin{matrix}m_{CO_2}=44\left(0,2+0,2\right)=17,6\left(g\right)\\m_{H_2O}=\left(0,4+0,2\right).18=10,8\left(g\right)\end{matrix}\right.\)
Xét \(\Delta m=m_{CO_2}+m_{H_2O}-m_{CaCO_3}=17,6+10,8-40=-11,6\left(g\right)\)
=> Khối lượng dd giảm 11,6 gam
PTHH: \(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\)
Ta có: \(n_{CO_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)=n_{CH_4}\)
Đặt \(\left\{{}\begin{matrix}n_{C_2H_4}=a\left(mol\right)\\n_{C_2H_2}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow a+b=\dfrac{5,04}{22,4}-0,075=0,15\) (1)
PTHH: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
Theo PTHH: \(28a+26b=4,1\) (2)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=n_{C_2H_4}=0,1\left(mol\right)\\b=n_{C_2H_2}=0,05\left(mol\right)\end{matrix}\right.\)
Mặt khác: \(n_{hh}=\dfrac{5,04}{22,4}=0,225\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,075}{0,225}\cdot100\%\approx33,33\%\\\%V_{C_2H_4}=\dfrac{0,1}{0,225}\cdot100\%\approx44,44\%\\\%V_{C_2H_2}=22,23\%\end{matrix}\right.\)
\(n_{CO_2}=\dfrac{8,96}{22,4}=0,4mol\)
\(m_{tăng}=m_{Br_2}=m_{C_2H_2}=2,6g\)
\(\Rightarrow n_{C_2H_2}=\dfrac{2,6}{26}=0,1mol\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
0,1 0,1
\(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
\(C_2H_2+\dfrac{5}{2}O_2\underrightarrow{t^o}2CO_2+H_2O\)
0,1 0,25 0,2
\(\Rightarrow n_{CO_2\left(CH_4\right)}=0,4-0,2=0,2mol\)
\(\Rightarrow n_{CH_4}=0,2mol\Rightarrow n_{O_2}=0,4mol\)
a)\(\%V_{CH_4}=\dfrac{0,2}{0,4}\cdot100\%=50\%\)
\(\%V_{C_2H_2}=100\%-50\%=50\%\)
b)\(\Sigma n_{O_2}=0,4+0,25=0,65mol\)
\(\Rightarrow V_{O_2}=0,65\cdot22,4=14,56l\)
\(\Rightarrow V_{kk}=14,56\cdot5=72,8l\)
nBr2 = 0,125 mol
Khi đốt cháy nA = 0,25 mol
mNaOH ban đầu = 36g => nNaOH = 0,9 mol
Gọi x, y lần lượt là số mol của CO2 và H2O
mdd = 180 + 44x + 18y
Vì NaOH dư do đó chỉ tạo muối trung hòa
CO2 +2NaOH → Na2CO3 + H2O
x 2x
nNaOH dư = 0,9 – 2x
có 2 , 75 % = 40 ( 0 , 9 - 2 x ) 180 + 44 x + 18 y . 100 % (1)
=> 81,21x + 0,495y = 31,05n
2,8 lít khí A tác dụng với 0,125 mol Br2
=> 5,6 lít khí A tác dụng với 0,25 mol Br2
Gọi số mol khí của CH4, C2H4 và C2H2 lần lượt là a; b; c
Ta có a + b + c = 0,25 mol
Và b + 2c = 0,25
=> a = c
=> khi đốt cháy hỗn hợp A cho nCO2 = nH2O
Thay vào (1) => x = y = 0,38 mol
Bảo toàn C, H khi đốt cháy ta có
=> %VCH4 = %VC2H2 = 48%
%VC2H4 = 4%
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 )
\(m_{CaCO_3}=0,8.100=80g\)