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Un amestec de clorura de metil si diclormetan contine 58% Cl. Determinati raportul masic al celor 2 componenti halogenati.

Răspuns :

MCH3Cl=50,5g/mol
MCH2Cl2=85g/mol

50,5x+85y.................35,5x+71y
100................................58

2929x+4930y=3550x+7100y=>621x=2170=>x/y=7/2

m CH3Cl=7*50,5=353,5g

m CH2Cl2=2*85=170g

x/y=353,5/170=2/1

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