unsure on how to use equation for this

Question

Given the following equation: 4Na(s) + O2(g ) → 2Na2O(s)

How many L of O2 gas at STP, are needed to react with 15.0 g of Na?

Answer 1

Great Stoich Problem!

Calculate mols of Na reacted = 15.0 g / 23g/ mol Na = .652 mol Na reacts.

Calculate mols of O2 required to react with .652 mol of Na using the balanced chemical equation.

.652 mol Na ( 1 mol O2 / 4 mol Na ) = .163 mol O2

Use ideal gas law to convert mols of O2 to liters.

STP means standard temperature and pressure

PV =nRT

V = nRT/P = .163 mol O2 ( .0821atm lit/ mol K) 273 K / 1 atm = 3.65 liters of O2 required to react with 15.0 g of Na at STP conditions.

Yours in Chemistry Success!

Dale Swanson

Answer 2

So a couple of important reminders that you should always remember is what STP is and how many liters a single mole of gas occupies. STP stands for standard pressure and temperature, which pressure is 1 atm and temperature is 273 Kelvins. At STP a single mole of gas occupies 22.4Liters which is an important conversion tool.

So we have the reaction and we want to find how many liters of oxygen gas at STP does are needed to react with 15.0 grams of Na. The equation is balanced so you don't have to worry about that. We need to convert from Na to oxygen using the reaction but our starting units of Na has to be in moles. We have grams so that last thing we need to solve this problem is molar mass. Na's molar mass is 23 grams/mole. So we convert from grams to moles of Na and then convert from moles of Na to moles of Oxygen and then we convert to liters as shown below:

15.0 grams of Na × (1 mole of Na/23 grams of Na) = 0.652 moles of Na

0.652 moles of Na × (1 mole of O2(g)/4 moles of Na(s)) = 0.163 moles of O2

0.163 moles of O2(g) ×(22.4 liters of gas/1 mole of gas) = 3.65214 Liters of O2

The answer is 3.65214 Liters of O2 is needed to react with 15 grams of Na.

If you got anymore questions just away.