Problem 5-81

CH₃OH(ℓ) + O₂(g) → CO₂(g) + H₂O(g)

First, balance the equation

Then, the number of mole of CH₃OH is:

50.0 mL CH3OH ×	0.850 g	= 42.5 g
	mL

42.5 g ×	1 mol	= 1.326 mol CH3OH
	32.042 g

The number of mole of oxygen present in 22.8 L O₂(g) at 27 °C and 2.00 atm is, using PV=nRT: n = 1.851 mol O₂

Once the equation is balanced, it is quite easy to identify the limiting reactant.
Then, the number of mole of H₂O produced, assuming a %yield = 100% can be calculated