Question

179,2 L de CH4 conduc prin clorurare la CH3Cl, CH2Cl2 si CH4 nereactionat in raport molar 4:3:1. Cantitatea de CH3Cl este:

Answer 1

179.2 L de CH4 = 179.2L/22.4 L/mol =>8 moli de CH4
4x+3x+x=8x
8x=8 => x=1

CH4 + Cl2 -> CH3Cl (total 4 moli)
CH4+ 2Cl2 -> CH2Cl2(total 3 moli)
CH4 -> CH4 (total 1 mol)

deci cantitatea CH3Cl este 4 moli = 4*MCH3Cl = 4*(12+3+35.5)