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PTHH: \(\left\{{}\begin{matrix}C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\\2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\end{matrix}\right.\)
\(n_X=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
Bảo toàn C: \(n_{CO_2}=n_C=2n_X=2.0,25=0,5\left(mol\right)\)
PTHH: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\)
0,5--------------------->0,5
=> mCaCO3 = 0,5.100 = 50 (g)
Theo CTHH: \(n_C=2n_{hh}=\dfrac{2,24}{22,4}.2=0,2\left(mol\right)\)
PTHH:
\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\\ C_2H_6O+3O_2\underrightarrow{t^o}2CO_2+3H_2O\\ 2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)
Bảo toàn C: \(n_{CO_2}=n_C=0,2\left(mol\right)\)
PTHH: \(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3\downarrow+H_2O\)
0,2----->0,2
=> mkết tủa = 0,2.100 = 20 (g)
\(n_{hhkhí}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\\ m_{tăng}=m_{C_2H_4}=4,2\left(g\right)\\ n_{C_2H_4}=\dfrac{4,2}{28}=0,15\left(mol\right)\\ n_{CH_4}=0,35-0,15=0,2\left(mol\right)\\ \left\{{}\begin{matrix}\%V_{C_2H_4}=\dfrac{0,15}{0,35}=42,85\%\\\%V_{CH_4}=100\%-42,85\%=57,15\%\end{matrix}\right.\)
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
\(m_{CaCO_3}=0,5.100=50\left(g\right)\)
a, \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
\(n_{Br_2}=\dfrac{16}{160}=0,1\left(mol\right)\)
Theo PT: \(n_{C_2H_2}=\dfrac{1}{2}n_{Br_2}=0,05\left(mol\right)\)
\(\Rightarrow V_{C_2H_2}=0,05.22,4=1,12\left(l\right)\)
\(\Rightarrow V_{CH_4}=3,36-1,12=2,24\left(l\right)\)
b, \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
\(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)
\(n_{CH_4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT: \(n_{O_2}=2n_{CH_4}+\dfrac{5}{2}n_{C_2H_2}=0,325\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,325.22,4=7,28\left(l\right)\Rightarrow V_{kk}=5V_{O_2}=36,4\left(l\right)\)
Theo PT: \(n_{CO_2}=n_{CH_4}+2n_{C_2H_2}=0,2\left(mol\right)\)
\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)
\(\Rightarrow n_{CaCO_3}=n_{CO_2}=0,2\left(mol\right)\Rightarrow m_{CaCO_3}=0,2.100=20\left(g\right)\)
Gọi CTHH của X là: CxHy
Theo đề, ta có:
\(d_{\dfrac{X}{H_2}}=\dfrac{M_{C_xH_y}}{M_{H_2}}=\dfrac{M_{C_xH_y}}{2}=15\left(lần\right)\)
=> \(M_{C_xH_y}=30\left(g\right)\)
a. PTHH: \(4C_xH_y+\left(4x+y\right)O_2\overset{t^o}{--->}4xCO_2+2yH_2O\) (1)
\(CO_2+Ca\left(OH\right)_2--->CaCO_3\downarrow+H_2O\) (2)
Ta có: \(n_{CaCO_3}=\dfrac{30}{100}=0,3\left(mol\right)\)
Theo PT(2): \(n_{CO_2}=n_{CaCO_3}=0,3\left(mol\right)\)
=> \(m_{C_{\left(CO_2\right)}}=m_{C_{\left(X\right)}}=0,3.12=3,6\left(g\right)\)
(Lỗi đề thì phải bn nhé.)
\(n_{hh}=\dfrac{11,2}{22,4}=0,5mol\)
Gọi \(\left\{{}\begin{matrix}n_{CH_4}=x\\n_{C_2H_4}=y\end{matrix}\right.\)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
x x ( mol )
\(C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\)
y 2y ( mol )
\(n_{CaCO_3}=\dfrac{80}{100}=0,8mol\)
\(Ca\left(OH\right)_2+CO_2\rightarrow\left(t^o\right)CaCO_3+H_2O\)
0,8 0,8 ( mol )
Ta có:
\(\left\{{}\begin{matrix}x+y=0,5\\x+2y=0,8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,3\end{matrix}\right.\)
\(\%V_{CH_4}=\dfrac{0,2}{0,5}.100=40\%\)
\(\%V_{C_2H_4}=100\%-40\%=60\%\)
\(m_{tăng}=m_{Ca\left(OH\right)_2}+m_{CaCO_3}=0,8.\left(74+100\right)=139,2g\)
V(C2H5OH) = 80.69% = 55,2 (ml)
m(C2H5OH) = 55,2.0,8 = 44,16 (g)
n(C2H5OH) = 44,16/46 = 0,96 (mol)
PTHH:
C2H5OH + 3O2 ---t°---> 2CO2 + 3H2O
Mol: 0,96 ---> 2,88 ---> 1,92
m(CO2) = 1,92.44 = 84,48 (g)
m(tăng) = m(CO2) = 84,48 (g)
\(n_{hh}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ M_{hh}=14.2=28\left(\dfrac{g}{mol}\right)\)
Áp dụng sơ đồ đường chéo:
\(\dfrac{n_{C_2H_2}}{n_{C_2H_4}}=\dfrac{V_{C_2H_2}}{V_{C_2H_4}}=\dfrac{28-28}{28-26}=\dfrac{0}{2}\)
Sai đề?