Question

For this reaction, 6.66 dollar ? methane (CH_4) reacts with 25.8 g capon tetrachloride methane (CH_4) (g) carbon ten chloride (g) -methane What is the maximum mass of dichloromethane (CH_2Cl_2) that can be famed0 g What is the FORMULA foil the limiting reagent? What mass of the excess reagent remains after the reaction is complete? g

Answer 1

The equation for the reaction is: CH_4 + Ccl_4 rightarrow 2CH_2cl_2 Mass of methane = 6.66g Moles of methane = 6.66g times 1 mole/16.04g = 0.415 moles Moles of CCl_4 = 25.8 g times 1 mole/153.82 g = 0.168 moles From the balanced equation, 1 mole of CH_4 reacts with 1 mole of CCl_4 So ccl_4 is the limiting reactants. 0.168 mole of CH_4 reacts with 0.168 moles of ccl_4 to give 2 times 0.168 = 0.33 moles of CH_2cl_2 Mass of product - CH_2cl_2 = 0.336 moles times 85.03 g = 28.6 g Formula of limiting reagents is ccl_4 methane is excess Mass of methane reacted = 0.168 moles times 16.04 g/1 mole = 2.69 g Mass of excess methane = 6.66g - 2.69 g = 3.97g