Answer:
Explanation:
2Câ‚‚Hâ‚‚(l) Â + Â 50â‚‚(g) = Â 4COâ‚‚(g) Â + Â 2Hâ‚‚O(g)
2 mol      5 mol
2 mol of Câ‚‚Hâ‚‚ reacts with 5 mol of Oâ‚‚
37 mol of Câ‚‚Hâ‚‚ Â will require 5/2 x 37 mol of Oâ‚‚
= 92.5 mol of Oâ‚‚
But oxygen available is only 81 mol so formation of product will be controlled by oxygen available . Hence oxygen is the limiting factor.