\(C_2H_2+\dfrac{5}{2}O_2\underrightarrow{t^o}2CO_2+H_2O\)
x------>2,5x---->2x---->x
\(2x+x=\dfrac{2,7099.10^{23}}{6,02.10^{23}}=0,45\left(mol\right)\Rightarrow x=0,15\left(mol\right)\)
\(N_{CO_2}=2.0,15.6,02.10^{23}=1,806.10^{23}\left(phân.tử\right)\\ N_{H_2O}=0,15.6,02.10^{23}=9,03.10^{22}\left(phân.tử\right)\)