Consider the mixture of propane, C3H8, and O2 shown here. (c) How many molecules of CO2, H2O, C3H8, and O2 will be present if the reaction goes to completion?

Video transcript

Hello everyone today. We are being given the following image and it shows a mixture of nitrogen and oxygen gasses at room temperature. Now, when the temperature of the mixture is raised nitric oxide and nitric dioxide or nitrogen dioxide are produced. And we have the following chemical equation here. We are then asked to find the number of molecules of each substance present when the reaction is complete. And so we see here that we have a total of five nitrogen molecules As well as a six oxygen molecules here. And so to start, we have to find the limiting reactant. And so to do that, we must find the molecules of nitric oxide and nitrogen dioxide for each reactant. So we'll start off with into so we'll start off with our into our nitrogen gas. So we're going to take how many moles that we have of that which is five moles of nitrogen gas. And we're gonna multiply that by the multiple ratio. And we do that using the chemical equation provided to us, we see that for every two moles of nitrogen gas and the dominator here, we have two moles of nitric oxide present. Our units are going to cancel out and this is going to leave us with five moles of nitric oxide. And to find the moles of nitrogen dioxide, we have to use the same reactant. So five moles of our nitrogen gas and they're going to multiply that by the multiple ratio. So once again we have two moles of nitrogen gas that produced two moles of nitrogen dioxide. Once again, we are going to be left with five moles of nitrogen dioxide for oxygen, we're gonna go through the same process. So we're gonna take our six moles of 02 guests that we have. We're gonna multiply that by the multiple ratio that we have. And so for every three moles Of oxygen gas present, we have two moles of nitric oxide gas. And this is going to yield us four moles of nitric oxide for our nitrogen dioxide. We take our six moles of 02 once again and we multiply by our multiple ratio for every three moles of oxygen gas, we have two moles of nitrogen dioxide giving us four moles of N. 02. Now we see that oxygen actually produces the least amount of products for each versus five. And so oxygen gas is our limiting reactant. So let me react in L. R. And so to find the number or the amount of nitrogen remaining, we must find out how much reacted. So we then have to take our six moles of 0. 2. And to cancel out moles of oxygen, we have to use our multiple ratio of nitrogen gas to oxygen gas. And we see that we perform the experiment with two moles of nitrogen gas as well as three moles of oxygen gas. And this is going to give us four moles of nitrogen gas when our units cancel out to find the nitrogen gas remaining. We must take that five moles that we initially had. We must have attracted by our molds that we produced when reacting with oxygen to give us one mole of nitrogen gas remaining. And so in the very end, we're going to have one mole of this nitrogen gas. We're going to have zero moles of oxygen because we've used it all. So we'll say we'll have zero molecules of 02. We're going to have one molecule of into. We're then going to have four molecules of both nitric oxide and nitrogen dioxide. And so this is going to be our final answer. I hope this helped. And until next time.