$n_{O_2}=\dfrac{8,46}{22,4}≈0,378mol$
$PTHH :$
$CH_4+2O_2\overset{t^o}\to CO_2+2H_2O$
a.Theo pt :
$n_{CH_4}=\dfrac{1}{2}.n_{O_2}=\dfrac{1}{2}.0,378=0,189mol$
$⇒m_{CH_4}=0,189.16=3,024g$
b.Theo pt :
$n_{CO_2}=\dfrac{1}{2}.n_{O_2}=\dfrac{1}{2}.0,378=0,189mol$
$⇒V_{CO_2}=0,189.22,4=4,2336l$
c.Cách 1 :
Theo pt :
$n_{H_2O}=n_{O_2}=0,378mol$
$⇒m_{H_2O}=0,378.18=6,804g$
Cách 2 :
$m_{O_2}=0,378.32=12,096g$
$m_{CO_2}=0,189.44=8,316g$
Áp dụng định luật bảo toàn khối lượng , ta có :
$m_{H_2O}=m_{CH_4}+m_{O_2}-m_{CO_2}=3,024+12,096-8,316=6,804g$