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\(2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\\ n_{C_2H_2}=1\left(mol\right)\Rightarrow n_{O_2}=\dfrac{5}{2}.1=2,5\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=2,5.22,4=56\left(l\right)\\ Vì:V_{O_2}=20\%V_{kk}\\ \Rightarrow V_{kk}=\dfrac{100}{20}.V_{O_2}=5.56=280\left(l\right)\)
\(n_{C_2H_2}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\\ a,2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\\ b,n_{CO_2}=0,125.2=0,25\left(mol\right)\\ m_{CO_2}=0,25.44=11\left(g\right)\\ c,n_{O_2}=\dfrac{5}{2}.0,125=0,3125\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,3125.22,4=7\left(l\right)\\ \Rightarrow V_{kk\left(đktc\right)}=\dfrac{100}{20}.7=35\left(l\right)\)
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: \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
\(2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\\ n_{C_2H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ n_{CO_2}=2.0,25=0,5\left(mol\right)\\ a,V_{CO_2\left(đktc\right)}=0,5.22,4=11,2\left(l\right)\\ b,n_{O_2}=\dfrac{5}{2}.0,25=0,625\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,625.22,4=14\left(l\right)\\ V_{kk\left(đkct\right)}=\dfrac{100}{20}.14=70\left(lít\right)\)
PT: \(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)
Theo PT: \(n_{O_2}=\dfrac{5}{2}n_{C_2H_2}=2,5\left(mol\right)\)
\(\Rightarrow V_{O_2}=2,5.22,4=56\left(l\right)\Rightarrow V_{kk}=\dfrac{V_{O_2}}{20\%}=280\left(l\right)\)
\(n_{C_2H_2}=\dfrac{5,6}{22,4}=0,25mol\)
\(C_2H_2+\dfrac{5}{2}O_2\underrightarrow{t^o}2CO_2+H_2O\)
0,25 0,625
\(V_{O_2}=0,625\cdot22,4=14l\)
\(V_{kk}=5V_{O_2}=5\cdot14=70l\)
\(n_{C_2H_2}=\dfrac{5,6}{22,4}=0,25mol\)
\(2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\)
0,25 0,625 ( mol )
\(V_{O_2}=0,625.22,4=14l\)
\(V_{kk}=V_{O_2}.5=14.5=70l\)
PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
Ta có: \(n_{CH_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
a, Theo PT: \(n_{O_2}=2n_{CH_4}=1\left(mol\right)\)
\(\Rightarrow V_{O_2}=1.22,4=22,4\left(l\right)\)
b, \(V_{kk}=\dfrac{22,4}{20\%}=112\left(l\right)\)
Đáp án: B