Câu 6: C
\(n_{CO_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
\(n_{H_2O}=\dfrac{14,4}{18}=0,8\left(mol\right)\)
=> \(\left\{{}\begin{matrix}n_C=0,6\left(mol\right)\\n_H=1,6\left(mol\right)\end{matrix}\right.\)
=> \(n_O=\dfrac{12-0,6.12-1,6.1}{16}=0,2\left(mol\right)\)
Xét nC : nH: nO = 0,6 : 1,6 : 0,2 = 3 : 8 : 1
=> CTPT: (C3H8O)n
Mà \(M_X=\dfrac{1,8}{\dfrac{0,96}{32}}=60\left(g/mol\right)\)
=> n = 1
=> CTPT:C3H8O
Câu 7:
a) \(n_{H_2O}=\dfrac{9}{18}=0,5\left(mol\right)\)
\(n_{CO_2}=\dfrac{22}{44}=0,5\left(mol\right)\)
=> \(\left\{{}\begin{matrix}n_C=0,5\left(mol\right)\\n_H=1\left(mol\right)\end{matrix}\right.\)
Xét mC + mH = 0,5.12 + 1.1 = 7 (g)
=> A không chứa O
b)
Xét nC : nH = 0,5 : 1 = 1 : 2
=> CTPT: (CH2)n
Mà \(M_A=7.4=28\left(g/mol\right)\)
=> n = 2
=> CTPT: C2H4
