Question

Answer 1

\(n_{C_2H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: \(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)

             0,15------------>0,3-->0,15

=> \(\left\{{}\begin{matrix}n_{CO_2}=0,3\left(mol\right)\\n_{H_2O}=0,15\left(mol\right)\end{matrix}\right.\)

Answer 2

\(PTHH:2C_2H_2+5O_2-nhiệtđộ->4CO_2+2H_2O\)

Ta có: \(n_{C_2H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

=> \(n_{CO_2}=2.n_{C_2H_2}=2.0,15=0,3\left(mol\right)\\ =>n_{H_2O}=n_{C_2H_2}=0,15\left(mol\right)\)