Đáp án:
Giải thích các bước giải:
$a,CH_4$ $l=không$ $tác$ $dụng.$
$PTPƯ:C_2H_4+Br_2→C_2H_4Br_2$
$b,n_{C_2H_4Br_2}=\dfrac{4,9}{188}=0,026mol.$
$Theo$ $pt:$ $n_{C_2H_4}=n_{C_2H_4Br_2}=0,026mol.$
$⇒V_{C_2H_4}=0,026.22,4=0,5824l.$
$⇒V_{CH_4}=2,8-0,5824=2,2176l.$
$c,PTPƯ:CH_4+2O_2\overset{t^o}\to$ $CO_2+2H_2O$ $(1)$
$C_2H_4+3O_2\overset{t^o}\to$ $2CO_2+2H_2O$ $(2)$
$n_{hh khí}=\dfrac{2,8}{22,4}=0,125mol.$
$⇒n_{CH_4}=0,125-0,026=0,099mol.$
$Theo$ $pt1:$ $n_{O_2}=2n_{CH_4}=0,198mol.$
$Theo$ $pt2:$ $n_{O_2}=3n_{C_2H_4}=0,078mol.$
$⇒n_{O_2}(2pt)=0,198+0,078=0,276mol.$
$⇒V_{O_2}=0,276.22,4=6,1824l.$
$⇒V_{kk}=6,1824.5=30,912l.$
chúc bạn học tốt!