methyl ethanoate

09. oktober 2009 af msn88 (Slettet)

hej

der står i min opg at

methyl ethanoate is produced from acid and methanol

CH3COOH + CH3OH --> CH3COOCH3 + H20

we want 250 g methyl ethanoate the percent yield is expected to be 60 %

a. how many ml 96 w/w% acetic is needed

b. how many ml 100 w/w% methanol is nedded when we want an excess of 30 %

hvilken formel skal jeg bruge

250 g = m

CH3COOCH3 + H20 = molarmassen af den ? mol/L

Brugbart svar (0)

Svar #1
10. oktober 2009 af mathon

se
https://www.studieportalen.dk/Forums/Thread.aspx?id=744643

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