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\(n_{CO_2}=\frac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH : \(C+O_2\underrightarrow{t^0}CO_2\)
PT : 1mol 1mol
Đề : 0,4mol ?mol
=> \(n_{O_2}=\frac{0,4\cdot1}{1}=0,4\left(mol\right)\)
=> \(V_{O_2}=0,4\cdot22,4=8,96\left(l\right)\)
\(V_{kk}\cdot20\%=V_{O_2}\Rightarrow V_{kk}=\frac{V_{O_2}}{20\%}=\frac{8,96}{20\%}=44,8\left(l\right)\)
=> \(V_{kk}=44,8l\)
a) PTHH: 2C2H2 + 5O2 --to--> 4CO2 + 2H2O
=> \(V_{O_2}=\dfrac{5}{2}V_{C_2H_2}=\dfrac{5}{2}.4=10\left(l\right)\)
b) \(\left\{{}\begin{matrix}n_{C_2H_2}=\dfrac{3,9}{26}=0,15\left(mol\right)\\n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\end{matrix}\right.\)
LTL: \(\dfrac{0,15}{2}< \dfrac{0,4}{5}\) => O2 dư
=> \(\left\{{}\begin{matrix}n_{CO_2}=0,15.2=0,3\left(mol\right)\\n_{O_2\left(pư\right)}=\dfrac{5}{2}.0,15=0,375\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow M_{hh}=\dfrac{0,3.44+\left(0,4-0,375\right).32}{0,3+0,4-0,375}=\dfrac{560}{13}\left(\dfrac{g}{mol}\right)\)
=> dhh/H2 = \(\dfrac{\dfrac{560}{13}}{2}=\dfrac{280}{13}\)
a, Ta có: \(n_{C_2H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PT: \(2C_2H_2+5O_2\underrightarrow{^{t^o}}4CO_2+2H_2O\)
\(n_{O_2}=\dfrac{5}{2}n_{C_2H_2}=0,5\left(mol\right)\Rightarrow V_{O_2}=0,5.22,4=11,2\left(l\right)\)
b, \(V_{kk}=\dfrac{V_{O_2}}{20\%}=56\left(l\right)\)
c, - Hiện tượng: Br2 nhạt màu dần.
PT: \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
\(a.2H_2+O_2-^{t^o}\rightarrow2H_2O\\ b.n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ n_{O_2}=\dfrac{1}{2}n_{H_2}=0,2\left(mol\right)\\ \Rightarrow V_{O_2}=0,2.22,4=4,48\left(l\right)\\ TrongkhôngkhíO_2chiếm20\%\\ \Rightarrow V_{kk}=\dfrac{4,48}{20\%}=22,4\left(l\right)\)
a, Ta có: \(n_{CO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)=n_C\)
\(n_{H_2O}=\dfrac{10,8}{18}=0,6\left(mol\right)\Rightarrow n_H=0,6.2=1,2\left(mol\right)\)
m = mC + mH = 0,4.12 + 1,2.1 = 6 (g)
b, Theo ĐLBT KL, có: m + mO2 = mCO2 + mH2O
⇒ mO2 = 22,4 (g) \(\Rightarrow n_{O_2}=\dfrac{22,4}{32}=0,7\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,7.22,4=15,68\left(g\right)\)
\(\Rightarrow V_{kk}=\dfrac{15,68}{20\%}=78,4\left(g\right)\)
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4mol\)
\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)
0,4 0,2 ( mol )
\(V_{O_2}=0,2.22,4=4,48l\)
\(V_{kk}=V_{O_2}.5=4,48.5=22,4l\)
Ta có: \(n_{C_2H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PT: \(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)
Theo PT: \(n_{O_2}=\dfrac{5}{2}n_{C_2H_2}=1\left(mol\right)\Rightarrow V_{O_2}=1.22,4=22,4\left(l\right)\)
Đáp án: A