368.92 g of HCOOH could be formed.
Explanation:
First we have to write the balanced equation as,
CH₃OH + O₂ → HCOOH + H₂O
Here the question said that 8.02 mol of CH₃OH and 15.49 mol of O₂.
In the above reaction methanol and oxygen is used in 1:1 ratio, but the moles are lesser in case of methanol, so CH₃OH is the limiting reagent.
Now by making use of 8.02 moles of methanol, we can produce 8.02 mol of HCOOH.
Molar mass of HCOOH is 46 g/mol.
So mass of HCOOH formed is moles ×molar mass of HCOOH.
8.02 moles × 46 g/mol = 368.92 g of HCOOH could be formed.
