How many liters of Oxygen gas (at STP) will be required to completely react with 6 grams of Sodium?

Answer 1

Recall that at STP, 1 mole of any gas = 22.4 L. With this in mind, we can approach the problem as follows:

Write the correctly balanced equation for the reaction taking place:

4Na + O₂ ==> 2Na₂O

Find moles of Na being used:

6 g Na x 1 mol Na / 23 g = 0.261 moles Na

From the balanced equation, find moles of O2 needed to react with 0.261 moles Na:

0.261 mols Na x 1 mol O₂ / 4 mols Na = 0.065 mols O₂ needed

Since it is at STP, 1 mol = 22.4 L so convert moles of O₂ to liters of O₂:

0.065 moles O₂ x 22.4 L / mol = 1.5 L of O₂

If you want to adjust the answer to the proper number of sig.figs., the answer would be 2 L of O₂