\(a.n_{CO_2}=\dfrac{3,36}{22,4}=0,15mol\\ CH_4+2O_2\xrightarrow[]{t^0}CO_2+2H_2O\\ n_{CH_4}=n_{CO_2}=0,15mol\\ V_{CH_4}=0,15.22,4=3,36l\\ b.n_{O_2}=2n_{CO_2}=0,3mol\\ V_{O_2}=0,3.22,4=6,72l\\ V_{KK}=6,72:\dfrac{1}{5}=33,6l\)
Đúng 2
Bình luận (0)