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\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\\ n_{O_2}=\dfrac{n_{H_2}}{2}\Rightarrow V_{O_2\left(đktc\right)}=\dfrac{V_{H_{2\left(đktc\right)}}}{2}=\dfrac{2,24}{2}=1,12\left(l\right)\\ Chọn.A\)
PTHH: 2H2 + O2 -> (t°) 2H2O
Mol: 0,1 ---> 0,05
VO2 = 0,05 . 22,4 = 1,12 (l)
=> B
a) \(n_{CH_4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
0,05-->0,1------->0,05
2C2H2 + 5O2 --to--> 4CO2 + 2H2O
0,125<--0,3125<----0,25
=> \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,05}{0,05+0,125}.100\%=28,57\%\\\%V_{C_2H_2}=\dfrac{0,125}{0,05+0,125}.100\%=71,43\%\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,05.16}{0,05.16+0,125.26}.100\%=19,753\%\\\%m_{C_2H_2}=\dfrac{0,125.26}{0,05.16+0,125.26}.100\%=80,247\%\end{matrix}\right.\)
b) \(n_{O_2}=0,1+0,3125=0,4125\left(mol\right)\)
=> \(V_{O_2}=0,4125.22,4=9,24\left(l\right)\)
=> Vkk = 9,24.5 = 46,2 (l)
Theo gt ta có: $n_{C_4H_{10}}=0,3(mol)$
$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$
Ta có: $n_{O_2}=1,95(mol)\Rightarrow V_{O_2}=43,68(l)$
a, Theo giả thiết ta có: \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
\(4P+5O_2--t^o->2P_2O_5\)
Ta có: \(n_{O_2}=\dfrac{5}{4}.n_P=0,125\left(mol\right)\Rightarrow V_{O_2\left(đktc\right)}=0,125.22,4=2,8\left(l\right)\)
b, Theo giả thiết ta có: \(n_{CH_4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
\(CH_4+2O_2--t^o->CO_2+2H_2O\)
Ta có: \(n_{O_2}=2.n_{CH_4}=0,1\left(mol\right)\Rightarrow V_{O_2\left(đktc\right)}=2,24\left(l\right)\)
PTHH: \(2C_4H_{10}+13O_2\overset{t^o}{--->}8CO_2+10H_2O\)
Ta có: \(n_{C_4H_{10}}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{13}{2}.n_{C_4H_{10}}=\dfrac{13}{2}.0,05=0,325\left(mol\right)\)
=> \(V_{O_2}=0,325.22,4=7,28\left(lít\right)\)
Chọn A