Suppose that 2. liters of air at stp is used to burn 50 g of carbon to form co2, and that the gaseous product is adjusted to stp. What is the volume and the average molar mass of the resulting mixture?

Suppose that 22.4 liters of air at STP is used to burn 1.50 g of carbon to form CO2, and that the gaseous product is adjusted to STP. What is the volume and the average molar mass of the resulting mixture?

Part One is set up for you to answer without thinking, without calculator, without much of anything except a basic knowledge of chemistry.

The reaction, combustion of carbon:

C(s)+O2(g)→CO2(g)C(s)+O2(g)→CO2(g)

And you see that for every mole of O2O2 removed from the air, a mole of CO2CO2 is added, so the total number of moles of gases is constant. The problem wants us to assume the gases are all ideal, so the total number of moles —regardless of the identity of the gases — determines the volume. It’s still 22.4 L.

The next part takes a bit more work: stoichiometry.

As a starting point, I’m going to steal Joakim Silva’s composition-of-air data:

• 
  nitrogen = 78.084% (molar mass 28.02)
  
• 
  oxygen = 20.946% (molar mass 32.00)
  
• 
  argon = 0.934% (molar mass 39.95)
  
• 
  carbon dioxide = 0.033% (molar mass 44.01)
  

So 22.4 L of air (which is 1.00 mol of gases) contains 0.20946 mol of O2O2. And note that each of those percentages corresponds to a decimal fraction of the number of moles of that gas (because the total is 1.00 mol).

And our 1.50 g of carbon is 0.1249 mol.

Since CC and O2O2 react in a 1:1 mole ratio, we can see that carbon is the limiting reactant. We will therefor produce 0.1249 mol of CO2CO2 and consume 0.1249 mol of O2O2.

Leaving us, after the combustion, with:

• 
  nitrogen = 0.78084 mol (unchanged)(molar mass 28.02)
  
• 
  oxygen = 0.20946 - 0.1249 = 0.08456 mol (molar mass 32.00)
  
• 
  argon = 0.00934 mol (unchanged) (molar mass 39.95)
  
• 
  carbon dioxide = 0.00033 + 0.1249 = 0.1252 mol (molar mass 44.01)
  

And to find the weighted average molar mass of the mixture is also easy from here, since the sum of the moles = 1.00.

av molar mass=(0.78084×28.02)+(0.08456×32.00)+av molar mass=(0.78084×28.02)+(0.08456×32.00)+

(0.00934×39.95)+(0.1252×44.01)=30.47 g/mol