Consider the following balanced chemical equation: 4Na (s) + O2 (g) ----> 2Na2O (s) . Calculate the volume of O2 gas (in L) needed at STP to react with 89 g Na.

Answer 1

Since the conditions are at STP, 1 mole of gas (ideal) will occupy 22.4 liters. Find moles of Na present and then using the stoichiometry of the balanced e of oxygen and finally convert to liters.

moles Na present = 89 g x 1 mole/23 g = 3.9 moles Na

moles O₂ = 3.9 mol Na x 1 mol O₂/4 mol Na = 0.98 moles

volume = 0.98 mol x 22.4 L/mol = 22 L