Isooctane, C8H18, is the component of gasoline from which
the term octane rating derives.
(d) How many moles of air are necessary for the combustion
of 1 mol of isooctane, assuming that air is 21.0% O2 by
volume? What is the volume in liters of this air at STP?

Question

Isooctane, C8H18, is the component of gasoline from which 
the term octane rating derives. 
(d) How many moles of air are necessary for the combustion 
of 1 mol of isooctane, assuming that air is 21.0% O2 by 
volume? What is the volume in liters of this air at STP?

Accepted Answer

Welcome back everyone. We're told that ammonia can be produced from the reaction of hydrogen gas and nitrogen gas at 500 degrees Celsius with a presence of a catalyst for part one. We need to calculate the moles of air needed to react with 1.5 moles of hydrogen gas, given that the air is 78% nitrogen by volume. In part two, we need to determine the volume and leaders of this air, which occupies at STP. Our first step is to write out a bounced reaction. So according to the prompt, we have hydrogen gas reacting with nitrogen gas and this is going to produce ammonia as a product and ammonia is also a gas. So to balance things out because we should recognize that we have six molds of, sorry, rather two moles of nitrogen on the reactant side and just one on the product side. We're going to place a coefficient of two in front of our ammonia which will now give us six moles of hydrogen on the product side. So to bounce that out, we're going to place a coefficient of three in front of our hydrogen gas on the reactant side. And now we can say that we have a balanced equation now that we've written out our balanced equation. Our next step is to go into part one of our prompts to solve for the moles of air needed to react with our moles of hydrogen gas based on our volume of nitrogen gas in the air. And so we want to recall that our percent volume is equivalent to our percent of moles of our components, where we want to recognize that our mole percent. We can interpret as our mole fraction, so we can interpret our mole percent as our mole fraction of our nitrogen gas and so we can take that percent being 78% in the prompt of nitrogen gas and interpret it as 0.780 being a decimal. Now, from our coefficients of our balanced equation, we can see that three moles of and let's use red here. So we can see that. Sorry about that. Three moles of our hydrogen gas reacts With one mole of our nitrogen gas. So based on this association, we can say that From the prompt, we know that we have 1.5 moles of gas Making up our hydrogen gas. So we can say our 1.5 moles of hydrogen gas reacts With, since we know that we have a ratio of 3-1 for our nitrogen or hydrogen to nitrogen gas, we're going to take 1.5 divided by three. And that is how we can say 1.5 moles of hydrogen gas reacts with .5 moles of nitrogen gas. So relating this to our equation, we can say that for the .5 moles of nitrogen gas that is reacting with our hydrogen. We are setting that equal to the amount of nitrogen gas as a fraction, which we calculated to be point and sorry about that .780 moles of nitrogen which makes up the air. And we're just going to plug that in as a variable as just moles of our air. And so we want to calculate the moles of air for part one. So we're going to divide both sides by . So by .780 moles of N two so that cancels out and what we would get is our molds of air isolated now equal to a value of 0.641 moles. So this will be our first answer for part one of our prompt And now we need to go to part two. So for part two of our prompt were asked to determine the volume of leaders occupied at S. T. P. For the air. So we're going to recall our formula which relates pressure and volume to the most of our gas times R gas constant R times temperature in kelvin and isolating for volume. We would have our most of our gas times R gas constant. R times our temperature in kelvin divided by our pressure. So, plugging in what we know for our variables. We have our most of our gas which we just confirmed for air. Since that's what part two is asking about being .641 moles of air Multiplied by R. Gas Constant R. Which from our lecture we recall is 0.08206. Leaders times a T. M's divided by moles times kelvin multiplied by our temperature given in our prompt as 500°C. So we are going to need to convert that to Kelvin. So we have 500°C Added to 273.15. And that's going to give us our kelvin temperature of and actually just to make a correction here. I almost got caught in a trap. We want to pay attention that in part two of our prompt we're told that we have standard temperature and pressure meaning that when we're plugging in for temperature here, we want to recall that at STP temperature is always equal to 2 73.15 Kelvin. I'm sorry just to 73.15 Kelvin. So we'll plug that in as 2 73.15 Kelvin. And now we're going to divide by our standard pressure which at STP we should recall is equal to 1 80 M. So that would be in our denominator. Just one A. T. M. So now canceling out our units, let's get rid of a T. M's, let's get rid of moles and let's get rid of kelvin and we'll be left with leaders as our final unit and this is going to give us our volume Of air and leaders that is occupied at STP being 14.4 L. So this would be our second final answer. To complete this example 14.4 L of air and everything highlighted in yellow represents our two final answers, or parts one and two of our prompt. If you have any questions, please leave them down below and I'll see everyone in the next practice video.

Isooctane, C8H18, is the component of gasoline from which the term octane rating derives. (d) How many moles of air are necessary for the combustion of 1 mol of isooctane, assuming that air is 21.0% O2 by volume? What is the volume in liters of this air at STP?

Verified Solution

Key Concepts

Stoichiometry of Combustion Reactions

Composition of Air

Standard Temperature and Pressure (STP)