Question

For the following reaction, 2.45 grams of methane (CH4) are allowed to react with 27.7 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? grams What is the FORMULA for the limiting reagent? What mass of the excess reagent remains after the reaction is complete? grams

Answer 1

Answer:

The limiting reactant is CH₄

26.0g of CH₂Cl₂ is the maximum amount that can be formed

4.15g CCl₄ will remain

Explanation:

The reaction of methane, CH₄, with carbon tetrachloride, CCl₄ is:

CH₄ + CCl₄ → 2CH₂Cl₂

To find the maximum mass of dichloromethane that can be determined we need to find moles of methane and carbon tetrachloride:

Moles CH₄:

2.45g * (1mol / 16.04g) = 0.153 moles

Moles CCl₄:

27.7g * (1mol / 153.82g) = 0.180 moles

That means just 0.153 moles of CCl₄ will react until CH₄ is over.

The limiting reactant is CH₄

Assuming the whole 0.153 moles will react, the moles of CH₂Cl₂ will be:

0.153 moles CH₄ * (2 moles CH₂Cl₂ / 1 mole CH₄) = 0.306 moles of CH₂Cl₂

The mass is (Molar mass dichloromethane: 84.93g/mol):

0.306 moles of CH₂Cl₂  * (84.93g / mol) = 26.0g of CH₂Cl₂

The moles of CCl₄ that remain are:

0.180 moles - 0.153 moles = 0.027 moles

In grams:

0.027 moles * (153.82g / mol) = 4.15g CCl₄