Write a balanced equation for the combustion of liquid methanol in air, assuming H2O(g) as a product.

The balanced equation for the combustion of liquid methanol in air, assuming H2O(g) as a product is

CH₃OH(l) + O₂(g) → CO₂(g) + H₂O(g)

From the question,

We are to write a balanced equation for the combustion of liquid methanol in air.

The combustion of liquid methanol in air is the reaction between methanol (CH₃OH) and oxygen (O₂). The reaction yields carbon(IV) oxide and water.

Now, for the balanced equation for the combustion of liquid methanol in air

The balanced chemical equation is

CH₃OH(l) + O₂(g) → CO₂(g) + H₂O(g)

Hence, the balanced equation for the combustion of liquid methanol in air, assuming H2O(g) as a product is CH₃OH(l) + O₂(g) → CO₂(g) + H₂O(g)

Learn more here: link

2 CH₃OH  +  3 O₂  ⇒  2 CO₂  +  4 H₂O

Explanation:

Methanol is CH₃OH.  Oxygen is O₂.  A combustion produces CO₂ and H₂O.  Create an equation using this information and balance.

CH₃OH  +  O₂  ⇒  CO₂  +  H₂O

2 CH₃OH  +  3 O₂  ⇒  2 CO₂  +  4 H₂O

i believe the answer is 5,000