a, $C_2H_4+ Br_2 \to C_2H_4Br_2$

b, 

$n_{C_2H_4Br_2}= \frac{4,9}{188}= 0,026 mol= n_{C_2H_4}$

=> $V_{C_2H_4}= 0,026.22,4= 0,5824l$

$V_{CH_4}= 2,8-0,5824= 2,2176l$

c,

$n_{hh}= \frac{2,8}{22,4}= 0,125 mol$

=> $n_{CH_4}= 0,125-0,026= 0,099 mol$

$CH_4+ 2O_2 \buildrel{{t^o}}\over\to CO_2+ 2H_2O$

$C_2H_4+ 3O_2 \buildrel{{t^o}}\over\to 2CO_2+ 2H_2O$

=> $n_{O_2}= 2n_{CH_4}+ 3n_{C_2H_4}= 2.0,099+3.0,026= 0,276mol$

$\%O_2= 20\%$

=> $V_{kk}= 5V_{O_2}= 5.0,276.22,4= 30,912l$

Đáp án:

 

Giải thích các bước giải:

$a,CH_4$ $l=không$ $tác$ $dụng.$

$PTPƯ:C_2H_4+Br_2→C_2H_4Br_2$

$b,n_{C_2H_4Br_2}=\dfrac{4,9}{188}=0,026mol.$

$Theo$ $pt:$ $n_{C_2H_4}=n_{C_2H_4Br_2}=0,026mol.$

$⇒V_{C_2H_4}=0,026.22,4=0,5824l.$

$⇒V_{CH_4}=2,8-0,5824=2,2176l.$

$c,PTPƯ:CH_4+2O_2\overset{t^o}\to$ $CO_2+2H_2O$  $(1)$

               $C_2H_4+3O_2\overset{t^o}\to$ $2CO_2+2H_2O$  $(2)$

$n_{hh khí}=\dfrac{2,8}{22,4}=0,125mol.$

$⇒n_{CH_4}=0,125-0,026=0,099mol.$

$Theo$ $pt1:$ $n_{O_2}=2n_{CH_4}=0,198mol.$

$Theo$ $pt2:$ $n_{O_2}=3n_{C_2H_4}=0,078mol.$

$⇒n_{O_2}(2pt)=0,198+0,078=0,276mol.$

$⇒V_{O_2}=0,276.22,4=6,1824l.$

$⇒V_{kk}=6,1824.5=30,912l.$

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