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\(n_{CO_2}=\dfrac{4.4}{44}=0.1\left(mol\right)\)
\(CH_4+2O_2\underrightarrow{^{^{t^0}}}CO_2+2H_2O\)
\(0.1.......0.2........0.1..........0.2\)
\(m_{CH_4}=0.1\cdot16=1.6\left(g\right)\)
\(V_{H_2O}=0.2\cdot22.4=4.48\left(l\right)\)
\(V_{kk}=5V_{O_2}=5\cdot0.2\cdot22.4=22.4\left(l\right)\)
nCH4 = 3,36 : 22,4 = 0,15 (mol)
pthh : CH4 + 2O2 -t--> CO2 + 2H2O
0,15 0,3
=> VO2 = 0,3 . 22,4 = 6,72 (L)
ta có : VO2 = 20% Vkk => Vkk = VO2 : 20% = 6,72 : 20% = 33,6 (L)
\(n_{CH_4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ PTHH:CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\\ Mol:0,15\rightarrow0,3\\ \rightarrow\left\{{}\begin{matrix}V_{O_2}=0,3.22,4=6,72\left(l\right)\\V_{kk}=6,72.5=33,6\left(l\right)\end{matrix}\right.\)
nCH4 = 4,48/22,4 = 0,2 (mol)
PTHH: CH4 + 2O2 -> (t°) CO2 + 2H2O
Mol: 0,2 ---> 0,4
Vkk = 0,4 . 22,4 : 21% = 128/3 (l)
\(n_P=\dfrac{m}{M}=\dfrac{6,2}{31}=0,2\left(mol\right)\)
\(PTHH:4P+5O_2-^{t^o}>2P_2O_5\)
tỉ lệ 4 : 5 : 2
n(mol) 0,2---->0,25---->0,1
`V(O_2)=nxx24,79=0,25xx24,79=6,1975(l)`
`V(kk)=6,1975:1/5=30,9875(l)`
a) \(n_{CH_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
0,4---------------->0,4
=> \(V_{CO_2}=0,4.22,4=8,96\left(l\right)\)
b) \(n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Xét tỉ lệ \(\dfrac{0,4}{1}>\dfrac{0,4}{2}\) => CH4 dư, O2 hết
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
0,4-------->0,2
=> \(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)
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)
a)
\(n_P = \dfrac{62}{31} = 2(mol)\\ 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ n_{O_2} = \dfrac{5}{4}n_P = 2,5(mol)\\ V_{O_2} = 2,5.22,4 = 56(lít)\\ V_{không\ khí} = \dfrac{56}{20\%} = 280(lít)\)
b)
\(n_P = \dfrac{31}{31} = 1(mol) ; n_{O_2} = \dfrac{23}{32} = 0,71875(mol)\\ \dfrac{n_P}{4} = 0,25 > \dfrac{n_{O_2}}{5} = 0,14375 \to P\ dư\\ n_{P\ pư} = \dfrac{4}{5}n_{O_2} = 0,575(mol)\\ m_{P\ dư} = 31 - 0,575.31 = 13,175(gam)\\ n_{P_2O_5} = \dfrac{2}{5}n_{O_2} = 0,2875(mol) \Rightarrow m_{P_2O_5} = 0,2875.142=40,825(gam)\)
PT: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
Ta có: \(n_{CH_4}=\dfrac{30,9875}{24,79}=1,25\left(mol\right)\)
a, \(n_{H_2O}=2n_{CH_4}=2,5\left(mol\right)\) \(\Rightarrow m_{H_2O}=2,5.18=45\left(g\right)\)
b, \(n_{O_2}=2n_{CH_4}=2,5\left(mol\right)\) \(\Rightarrow V_{O_2}=2,5.24,79=61,975\left(l\right)\)
Mà: O2 chiếm 1/5 thể tích không khí.
\(\Rightarrow V_{kk}=5V_{O_2}=309,875\left(l\right)\)
cho mk xin lỗi chữ õi ở câu b là sai nha oxi chứ không phải mk bấm nhầm