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a) \(n_{C_2H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
PTHH: \(2C_2H_2+5O_2\xrightarrow[]{t^o}4CO_2+2H_2O\)
0,6----->1,5
b) \(V_{kk}=\dfrac{1,5.22,4}{20\%}=168\left(l\right)\)
\(n_{C_2H_2}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\\ a,2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\\ b,n_{CO_2}=0,125.2=0,25\left(mol\right)\\ m_{CO_2}=0,25.44=11\left(g\right)\\ c,n_{O_2}=\dfrac{5}{2}.0,125=0,3125\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,3125.22,4=7\left(l\right)\\ \Rightarrow V_{kk\left(đktc\right)}=\dfrac{100}{20}.7=35\left(l\right)\)
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: \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
a, \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
\(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)
Ta có: \(n_{CH_4}+n_{C_2H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\left(1\right)\)
\(n_{CO_2}=n_{CH_4}+2n_{C_2H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CH_4}=0,25\left(mol\right)\\n_{C_2H_2}=0,05\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,25.22,4}{6,72}.100\%\approx83,33\%\\\%V_{C_2H_2}\approx16,67\%\end{matrix}\right.\)
Theo PT: \(n_{O_2}=2n_{CH_4}+\dfrac{5}{2}n_{C_2H_2}=0,625\left(mol\right)\Rightarrow m_{O_2}=0,625.32=20\left(g\right)\)
PTHH: 2C2H2 + 5O2 --to--> 4CO2 + 2H2O
\(\dfrac{V_{O_2}}{V_{C_2H_2}}=\dfrac{n_{O_2}}{n_{C_2H_2}}=\dfrac{5}{2}\Rightarrow V_{O_2}=2,5.\dfrac{5}{2}=6,25\left(l\right)\)
=> \(V_{kk}=6,25.5=31,25\left(l\right)\)