\(n_{H_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\\ n_{O_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\)
PTHH: 2H2 + O2 --to--> 2H2O
LTL: \(\dfrac{1,5}{2}< 1,5\rightarrow O_2\) dư
Theo pt: \(\left\{{}\begin{matrix}n_{O_2\left(pư\right)}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}.1,5=0,75\left(mol\right)\\n_{H_2O}=n_{H_2}=1,5\left(mol\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}m_{O_2\left(dư\right)}=\left(1,5-0,75\right).32=24\left(g\right)\\V_{O_2}\left(1,5-0,75\right).22,4=16,8\left(l\right)\\m_{H_2O}=1,5.18=27\left(g\right)\end{matrix}\right.\)
\(n_{H_2}=n_{O_2}=\dfrac{33,6}{22,4}=1,5\left(MOL\right)\)
pthh: \(2H_2+O_2\underrightarrow{t^O}2H_2O\)
LTL : \(\dfrac{1,5}{2}< \dfrac{1,5}{1}\)
=> O2 dư , H2 hết
theo pthh: nH2O = nH2 = 1,5 (mol)
=> \(m_{H_2O}=1,5.18=27\left(g\right)\)