Question

Care este cantitatea de Cl2 necesara pentru a obtine 23,9 g amestec echimolecular clorura de metilen si CCl4 pornind de la metan?

Answer 1

MCH2Cl2=85g/mol si MCCl4=154g/mol
raport echimolecular => x:x
m=n*M => mamestec=x(85+154)=239x=23,9 => x=0,1moli
............71g......1mol
CH4 + 2Cl2 -> CH2Cl2 + 2HCl
............7,1g.....0,1moli
...........142g....1mol
CH4 + 4Cl2 -> CCl4 + 4HCl
...........14,2g....0,1moli
mCl2=21,3g