The source of oxygen that drives the internal combustion
engine i...

Question

The source of oxygen that drives the internal combustion
engine in an automobile is air. Air is a mixture of gases, principally
N2179%2 and O2120%2. In the cylinder of an automobile
engine, nitrogen can react with oxygen to produce
nitric oxide gas, NO. As NO is emitted from the tailpipe of
the car, it can react with more oxygen to produce nitrogen
dioxide gas. (c) The production of NOx gases is an unwanted
side reaction of the main engine combustion process
that turns octane, C8H18, into CO2 and water. If 85% of
the oxygen in an engine is used to combust octane and the
remainder used to produce nitrogen dioxide, calculate how
many grams of nitrogen dioxide would be produced during
the combustion of 500 g of octane.

Accepted Answer

Welcome back everyone in this example, we have nitrogen gas reacting with magnesium to form magnesium nitride at 800 degrees Celsius, we're told that the source of nitrogen in this reaction is atmospheric nitrogen from the air and air contains about 78% nitrogen, 20 1% oxygen and 1% other gasses. Since the reaction is exo thermic, the released heat can result in the reaction between nitrogen and oxygen to form nitric oxide. So this is a second reaction. If 95% of the nitrogen in the air reacts with magnesium to form magnesium nitride and the remainder reacts with oxygen to form nitric oxide. How many grams of nitric oxide would be produced when 45 g of magnesium reacts with the excess air. So what we need to recognize is that we have two different reactions here. And let's write out our first reaction where the prop tells us we have nitrogen gas reacting with magnesium to form our product magnesium nitride. Now we need to make sure that this is a balanced reaction. And right now we can see on our product side that we have a subscript of three next to our magnesium, meaning we need three moles of our magnesium for this to be balanced on our reaction side. So we're going to place a coefficient of three on our reactant side in front of magnesium and now we have a balanced reaction. Now this reaction is important because what we can do is use it to figure out and start geometry step, the total amount of nitrogen that is reacting with magnesium to form our product here. And if we figure out that total amount and based on the prompt, recognize that only 95% of this nitrogen is reacting with magnesium. We can divide that out from our total amount of nitrogen to get our massive nitrogen that's reacting so that we can see based on that 5% that's remaining of our nitrogen is reacting with oxygen to produce our nitric oxide. So let's go ahead and begin with our first step which is going to be finding our mass of nitrogen that reacts. So we're going to begin with the first piece of information from the prompt which is that we are reacting nitrogen with our 45 g of magnesium. Actually let's use the color black here. So we have 45g of magnesium and we're going to go from magnesium and grams to moles by recalling its molar mass on our periodic table where we see that for one mole of magnesium. So we'll say one mole of MG. We have a molar mass of 24.305 g of magnesium. This allows us to cancel our units of grams And now we're at most and can look at the molar ratio from our balanced equation between magnesium and our moles of our nitrogen. And so we can see from our balanced equation that we have coefficients of three in front of magnesium and a coefficient of one in front of nitrogen. So we have a 3 to 1 molar ratio and we'll plug that in. We have three moles of magnesium for one mole of nitrogen, we can cancel out moles of magnesium. Now we're at moles of nitrogen and we want to find our mass of nitrogen that is reacting. So we're going to multiply by our next conversion factor to go from moles of nitrogen to go two g of nitrogen. Where we're going to again get the molar mass of two atoms of nitrogen from the periodic table. And we'll see that for one mole of nitrogen we have a mass of 28. g. Again, nitrogen is a diatonic molecule and so we take its molar mass from the periodic table and multiply it by two for this value here. Now we can cancel out our units of moles of nitrogen. We're left with g which is what we want. And this is going to yield a mass of 17.29 27. And sorry, let's make sure this is clear. So 17.29 27 g of nitrogen that reacts. We're now going to focus on getting the total nitrogen in our reaction or we can say the total nitrogen in grams. And what we want to do is recognize that according to the prompt, we have 95% of the nitrogen in the air reacting. Now, if we know that we have 95%, we're going to look at this in terms of mass percent recall that are mass percent formula relates our percentage. So we have 95% of nitrogen according to the prompt and we're relating this to our mass of nitrogen that reacts over its total mass available. So total mass of nitrogen. What we want to solve for is our total mass of nitrogen so that we can figure out also in a stroke. Yama tree step, how much is reacting with oxygen to form our nitric oxide and grams. So we're going to solve for our denominator here by plugging in what we know. We know our numerator. We know our massive nitrogen that reacts to will have 95 which will actually just write this as a decimal Now for simplicity sake. So 950.95 of our nitrogen. Well say g of nitrogen is equal to our massive nitrogen that reacts which we just determined is 17.29 27 g of nitrogen. And we want to solve for our denominator which is our total mass of nitrogen. So we're going to multiply both sides by our total massive nitrogen. That's going to give us .95g of nitrogen, multiplied by the total mass of N two equal to on our right hand side. We have 17.29 27 g of nitrogen that we know reacts And now we're going to isolate by dividing both sides by 270.95 g that cancels this on the left hand side. And now we have totally isolated our total mass of nitrogen, Which we can say is equal to the value of 18.20-8 g of nitrogen. So now that we have this value, we're going to write out our second reaction. We know that nitrogen is reacting with oxygen now which is also a diatonic molecule to produce nitric oxide, n. o. And now we want to make this bounce by placing a coefficient of two in front of our nitrogen monoxide. Now to add our units, these are both gasses and this is also a gaseous product. So now that we have our bounced reaction, we want to find our store geometry where we're going to figure out the massive nitric oxide produced from our nitrogen reacting with oxygen here. So we're going to begin with the info from the prompt which tells us that we have 5% of nitrogen reacting with oxygen. And so we can say that because we know our total mass of nitrogen is 18.20 to 8 g of nitrogen. We're going to multiply it by this 5%. So by 0.5 and sorry, that should be 0.5 for 5%. And this is going to give us our massive nitrogen reacting in the second reaction with oxygen where we have .91014g of nitrogen that reacts With or with oxygen. So let's get into our geometry step. So we will begin with what we just calculated .91014g of nitrogen. We're gonna go from grams of nitrogen to moles of nitrogen by recalling from our periodic table are molar mass of our di atomic molecule of nitrogen and we'll see that for one mole of nitrogen we have a mass of 28 point oh two g of nitrogen. We can cancel out grams of nitrogen and focus on going from moles of nitrogen two moles Of our product nitrogen monoxide or nitric oxide. And we'll see from our balanced equation that we have a 2- molar ratio. So we have two moles of nitric oxide for one mole of nitrogen. This allows us to cancel our units of moles of nitrogen and focus on going from moles of nitric oxide to our final answer of grams of nitric oxide. And we can see from our periodic table, the molar mass of nitric oxide for one mole is equal to 30.1 g of nitric oxide. We can get rid of our units of moles were left with grams of our final unit of nitric oxide. And this is going to give us our final answer where we see we have 1.95 g of nitric oxide produced when nitrogen reacts with oxygen in the air. And so this is going to be our final answer to complete this example. I hope everything I reviewed was clear. If you have any questions, leave them down below and I'll see everyone in the next practice video.

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