What mass of sodium oxide is produced from 4.23 grams of sodium and 20 liters of oxygen gas?

A limiting reagent problem since amounts are given for BOTH reactants.

2Na + O2 ==> Na2O
mols Na = grams/molar mass
mol O2 (is this at STP?). Assuming it is mols O2 = L/22.4 = ?

Convert mols Na to mols Na2O using the coefficients in the balanced equation.
Do the same for O2.
It is likely the two values will not b the same which means one of them is wrong. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
Convert the smaller value to grams Na2O. g = mols x molar mass.

To determine the mass of sodium oxide produced, we need to calculate the limiting reactant first. The limiting reactant is the substance that is completely used up in a chemical reaction and determines the maximum amount of product that can be formed.

Let's start with the balanced chemical equation for the reaction:
4 Na + O2 -> 2 Na2O

From the equation, we can see that 4 moles of sodium (Na) react with 1 mole of oxygen gas (O2) to produce 2 moles of sodium oxide (Na2O).

Step 1: Convert the given mass of sodium (Na) to moles:
Using the molar mass of sodium (Na), which is 22.99 g/mol:
4.23 g Na * (1 mol Na / 22.99 g Na) = 0.184 mol Na

Step 2: Convert the given volume of oxygen gas (O2) to moles:
Using the ideal gas law, PV = nRT, and we know that at STP (standard temperature and pressure), 1 mole of any ideal gas occupies 22.4 liters:
20 L O2 * (1 mol O2 / 22.4 L O2) = 0.893 mol O2

Step 3: Determine the limiting reactant:
To find the limiting reactant, compare the mole ratio of the reactants. From the balanced equation, we know that 4 moles of Na react with 1 mole of O2. Therefore, 4 moles of Na would require (4/1) * 0.893 mol O2 = 3.572 mol O2.

Comparing this calculation to the moles of O2 given (0.893 mol O2), we see that the actual amount of O2 is less than the required amount for complete reaction with 4.23 g of sodium. Therefore, oxygen gas (O2) is the limiting reactant.

Step 4: Calculate the mass of sodium oxide (Na2O) produced:
Using the mole ratio from the balanced equation, we know that 1 mole of O2 produces 2 moles of Na2O. Therefore:
0.893 mol O2 * (2 mol Na2O / 1 mol O2) * (61.98 g Na2O / 1 mol Na2O) = 109.55 g Na2O

Therefore, 109.55 grams of sodium oxide (Na2O) would be produced from 4.23 grams of sodium (Na) and 20 liters of oxygen gas (O2).