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clockTalking
@clockTalking
July 2022
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4,1 г смеси этилена и ацетилена способно прореагировать с 32 г Br2. Определите состав смеси
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alima1958
Verified answer
X--C2H2 4.1-x---C2H4
x
С2Н2+2Br2=C2H2Br4
4.1-x 32г
С2Н4 + Br2 = C2H4Br2 (4,1-х)*160=28*32 х=1,5г С2Н2
28г 160г m(C2H4)=4.1-1.5=2.6г
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Answers & Comments
Verified answer
X--C2H2 4.1-x---C2H4x
С2Н2+2Br2=C2H2Br4
4.1-x 32г
С2Н4 + Br2 = C2H4Br2 (4,1-х)*160=28*32 х=1,5г С2Н2
28г 160г m(C2H4)=4.1-1.5=2.6г