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\(n_C=\dfrac{3.6}{12}=0.3mol\)
\(C+O_2\underrightarrow{t^o}CO_2\)
0.3 0.3
\(V_{O_2}=0.3\times22.4=6.72l\)
\(V_{Kk}=6.72\times5=33.6l\)
\(n_{O_2}=\dfrac{44,8}{22,4}.20\%=0,4(mol)\)
Bảo toàn NT (O): \(n_{O_2}=n_{CO_2}=\dfrac{1}{2}n_{H_2O}\)
\(\Rightarrow n_{CO_2}=0,4(mol);n_{H_2O}=0,8(mol)\\ \Rightarrow V_{CO_2}=0,4.22,4=8,96(g);m_{H_2O}=0,8.18=14,4(g)\)
\(n_{O_2}=\dfrac{41,44}{22,4}.20\%=0,37(mol)\\ n_{H_2O}=\dfrac{4,68}{18}=0,26(mol)\)
Bảo toàn nguyên tố (O): \(n_{CO_2}=n_{O_2}=0,37(mol)\)
\(\Rightarrow V_{CO_2}=0,37.22,4=8,288(l)\)
BTKL: \(m_{hh}=m_{CO_2}+m_{H_2O}-m_{O_2}=0,37.44+4,68-0,37.32=9,12(g)\)
\(2C_2H_2+5O_2-^{t^o}\rightarrow4CO_2+2H_2O\\ n_{C_2H_2}=\dfrac{1,3}{26}=0,05\left(mol\right)\\ n_{O_2}=\dfrac{5}{2}n_{C_2H_2}=0,125\left(mol\right)\\ \Rightarrow V_{O_2}=2,8\left(l\right)\\ Vìtrongkhôngkhíchứa20\%O_2\\ \Rightarrow V_{kk}=\dfrac{2,8}{20\%}=14\left(l\right)\)
Đáp án A
Hỗn hợp khí gồm CH4, C2H6 và C3H8 là các hidrocacbon.
Mà khi đốt cháy 1 hidrocacbon bất kì ta luôn có:
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\(n_P=\dfrac{23,87}{31}=0,77\left(mol\right)\\ 4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\ n_{O_2}=\dfrac{5}{4}.0,77=0,9625\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,9625.22,4=21,56\left(l\right)\\ V_{kk\left(đktc\right)}=21,56.5=107,8\left(lít\right)\)
Ta có: \(n_P=\dfrac{23,87}{31}=0,77\left(mol\right)\)
PT: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)
Theo PT: \(n_{O_2}=\dfrac{5}{4}n_P=0,9625\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,9625.22,4=21,56\left(l\right)\)
\(\Rightarrow V_{kk}=\dfrac{V_{O_2}}{20\%}=107,8\left(l\right)\)