Calculate the volume of oxygen needed for complete combustion of ethene. Show your work.

Question

When ethene, C2H4 (g), is completely combusted, carbon dioxide, CO2(g), and water vapor, H2O(g), are produced according to the following balanced equation: 

C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (g)

If 11.2 L of carbon dioxide gas is produced at STP, answer questions a – c.(Assume that all volume measurements are made at the same temperature and pressure) 

Answer 1

Strategy: We can calculate the mole of O₂ if we know mole of CO₂ produced based on the balanced chemical equation. Once you know the moles of O₂, we can calculate the volume of O₂.

So, first we need to find how many moles of CO₂ are in 11.2 L of CO₂.

We know that at STP ==> 22.4 L of ANY gas = 1 mol of gas.

Therefore, 11.2 L of CO₂ at STP ==> (1 mol CO₂/22.4 L CO₂) * (11.2 L CO₂) = 0.500 mol CO₂ (3 Sig Fig)

According to the balanced equation, to produce 2 moles of CO₂, we need 3 moles of O₂

Therefore 0.500 mol of CO₂ would need ==> (3 mol O₂/2 mol CO₂) * (0.500 mol CO₂) = 0.750 mol O₂

Now we can convert mole of O₂ to Liters of O₂ using same conversion factor at STP.

Volume of O₂ = (22.4 L O₂/1 mol O₂) * (0.750 mol O₂) ==> 16.8 L of O₂