Question

For combustion of liquid methanol (CH3OH):

c. If you react 3.4 grams of methanol and 1.5 grams of oxygen gas, which is the limiting reactant?
d. How many grams of each product will form in the reaction?

Answer 1

Combustion of liquid methanol (CH₃OH) takes place in following way:

Every 2 mol of CH₃OH it requires 3mol of O₂ to form 3mol of CO₂ and 4 mol of H₂O

C. We get masses from the problem. Convert the mass to mol

molar mass of methanol= 32.04

molar mass of oxygen = 32

3.4g x (1mol CH₃OH/32.04g) = 0.106mol CH₃OH

1.5g x (1mol O₂ / 32g) = 0.047mol O₂

Now, we have to check if we have enough oxygen for combustion of methanol or not.

So, 0.106mol CH₃OH will require:

0.106mol CH₃OH x (3mol O₂/ 2mol CH₃OH)= 0.159 mol O₂

Since we have more moles of O₂ (0.159mol) than we require i.e. 0.047mol, we can say that methanol is the limiting reagent. All the methanol will get comnbusted before all the oxygen will react.

d. Hence, we can say that the reaction will require 0.159mol of O₂ and

0.159mol O₂ x (2mol CH₃OH/ 3mol O₂)= 0.106mol CH₃OH

This will produce:

0.106 mol of CH₃OH x (2mol CO₂/ 2mol CH₃OH)= 0.106mol of CO₂

0.106mol CO₂ x (44g/1mol CO₂)= 4.664g CO₂

0.106mol CH₃OH x (4mol H₂O/2mol CH₃OH)= 0.212 mol H₂O

0.212 mol H₂Ox (18.015g/1mol H₂O) =3.82g of H₂O