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a) \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: \(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\)
0,15<---0,3<----0,15
b) `m_{O_2} = 0,3.32 = 9,6 (g)`
c) `V_{CH_4} = 0,15.22,4 = 3,36 (l)`
a, PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
b, Ta có: \(n_{CH_4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{O_2}=n_{H_2O}=2n_{CH_4}=0,5\left(mol\right)\\n_{CO_2}=n_{CH_4}=0,25\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{CO_2}=0,25.44=11\left(g\right)\\m_{H_2O}=0,5.18=9\left(g\right)\end{matrix}\right.\)
c, Ta có: \(V_{O_2}=0,5.22,4=11,2\left(l\right)\)
Mà: VO2 = 1/5Vkk
\(\Rightarrow V_{kk}=11,2.5=56\left(l\right)\)
Bạn tham khảo nhé!
\(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\\ a,PTHH:4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\ b,n_{O_2}=\dfrac{5}{4}.0,4=0,5\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,5.22,4=11,2\left(l\right)\\ c,n_{P_2O_5}=\dfrac{2}{4}.0,4=0,2\left(mol\right)\\ m_{P_2O_5}=142.0,2=28,4\left(g\right)\)
nCH4 = 3,2/16 = 0,2 (mol)
PTHH: CH4 + 2O2 -> (t°) CO2 + 2H2O
Mol: 0,2 ---> 0,4
Vkk = 0,4 . 5 . 22,4 = 44,8 (l)
a, \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
b, \(n_{CH_4}=\dfrac{28}{22,4}=1,25\left(mol\right)\)
\(n_{CO_2}=n_{CH_4}=1,25\left(mol\right)\Rightarrow m_{CO_2}=1,25.44=55\left(g\right)\)
c, \(n_{O_2}=2n_{CH_4}=2,5\left(mol\right)\Rightarrow V_{O_2}=2,5.22,4=56\left(l\right)\)