PTHH : C2H4 +Br2 -----> C2H4Br2 Ta có ; nBr2=0.0175 (mol) ==> nC2H4=0.0175 (mol) ==>mC2h4=0.49(g) mà n(hỗn hợp khí)=3.36/22.4=0.15(mol) ==>nC2H6=0.15-0.0175=0.1325(mol) ==>mC2H6=3.975(g0 ==> m(hỗn hợp khí ) =4.465(g) ==> %mC2H4 =10.97% ==> %mC2H6= 89.03%
PTHH :
$C
_2H
_4 +Br
_2
\rig
ht
arrow C
_2H
_4Br
_2
$ Ta có
: $nBr
_2=0.0175 (mol)
\rig
ht
arrow nC
_2H
_4=0.0175 (mol)
\rig
ht
arrow mC
_2
H_4=0.49(g)
$ Mà
$n
_{hỗn hợp
}=
\frac{3
,36
}{22
,4
}=0.15(mol)
\rig
ht
arrow nC
_2H
_6=0.15-0.0175=0.1325(mol)
\rig
ht
arrow mC
_2H
_6=3.975(g
)$ $ \rig
ht
arrow m
_{hỗn hợp
} =4.465(g)
$ %
$m
_{C
_{2
}H
_{4
}} $ $=10.97
$%
$\rig
ht
arrow$ %
$m
_{C
_{2
}H
_{6
}}= 89.03
$%