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

Question

For the following reaction, 7.77 grams of methane (CH₄) are allowed to react with 27.4 grams of carbon tetrachloride .

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

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

What is the FORMULA for the limiting reagent?

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

Answer 1

Caitlyn

Atomic weights are : H, 1.008, C, 12.011 Cl 35.45

So 1 mole of methane has molecular weight 12.011 + 4 x 1.008 = 16.043

And 1 mole of carbon tetrachloride has molecular weight 12.011 + 4 x 35.45 = 153.811

So we have 7.77 / 16.043 = 0.48 moles of methane

27.4 / 153.811 = 0.18 moles of carbon tetrachloride

So the formula for the limiting reagent is CCl4 (carbon tetrachloride) as we have fewer moles of that then the methane.

We can only get 0.18 moles of dichloromethane (using the equation).

Molecular weight of this is 12.011 + 2 x 1.008 + 2 x 35.45 = 84.927

So mass is 0.18 x 84.927 = 15.29 g

0.48 - 0.18 = 0.3 moles of methane is left after the reaction, so 0.30 x 16.043 = 4.82 g

Mike

Answer 2

Write the correctly balanced equation for the reaction (note; it is different from the one given):

CH₄ + CCl₄ ==> 2CH₂Cl₂ ... balanced equation

moles CH₄ = 7.77 g x 1 mol/16.04 = 0.486 moles CH₄

moles CCl₄ = 27.4 g x 1 mol/154 = 0.178 moles CCl₄

Because the mol ratio in the correctly balanced equation of CH₄:CCl₄ is 1:1, we can see that CCl₄ is the limiting reactant.

To find the maximum mass of CH₂Cl₂ that can be formed, again we use the mol ratio of CH₂Cl₂ to CCl₄ which is 2:1 meaning that 1 mole of CCl₄ produces 2 moles of CH₂Cl₂.

0.178 mol CCl₄ x 2 mol CH₂Cl₂/mol CCl₄ = 0.356 mol CH₂Cl₂ x 85 g/mol = 30.3 g of CH₂Cl₂ (3 sig. figs.)

To find the mass of excess reagent (CH₄) left over, we know that of the 0.486 mol originally present, 0.178 moles were used (1:1 with CCl₄). Thus moles left over = 0.486 - 0.178 = 0.308 moles CH₄.

mass CH₄ = 0.308 moles x 16.0 g/mol = 4.93 g CH₄ left over (3 sig. figs.)