Question

For this reaction, 2.79 g methane (CH₄) reacts with 30.0 g carbon tetrachloride.

methane (CH₄) (g) + carbon tetrachloride (g) dichloromethane (CH₂Cl₂) (g)

What is the maximum mass of dichloromethane (CH₂Cl₂) that can be formed?  g

What is the FORMULA for the limiting reagent?

What mass of the excess reagent remains after the reaction is complete?  g

Answer 1

1. - 169.869 The reaction between methane and carbon tetrachloride is given by - CHylg) + CCl4 (9) 2 CHAų (g) Amol 1 mol 2 mol 16.048 153.82g 2x841939 16.049 153.82g so from above description it is clear that 16.04g of Che reacts completely with 183.82g of clue to form 169.86 g of cria . Mass of cle required to react with 2-79 g of the : :150842 x 2.799 = 2676 g of clue But given mass of ccl4 is a 30.og. so cale is present in excess and chly is the limiting reagent. .: Mars of et ų that can be prodcued = 46 709 x 2-799 . 16.04g 1 = 29.54g of CH₂. Formula of limiting reautant = Chap mass of excers reagent left = (3010 - 26+76)g. = 3.248