A hydrogen-filled balloon is ignited and 1.50 g of hydrogen is reacted with 12.0 g of oxygen. How many grams of water vapor form? (Assume that water vapor is the only product.)

Video transcript

All right. Hi, everyone. So this question says that a nitrogen filled balloon is ignited and 2.75 g of nitrogen reacts with 8.79 g of oxygen. How many grams of nitrogen oxide form? Assume that nitrogen oxide is the only product. And here we have four different answer choices labeled A through D proposing different masses for nitrogen oxide. So for this question, right, for this problem, we're going to have to first find the balanced chemical reaction for the process being described. So in terms of our reagents, we have nitrogen gas that's N two combining with oxygen gas or 02 reacting to yield nitrogen oxide or simply no, which is also a guess it's important to recognize that the chemical equation must be balanced before we can use it in dimensional analysis. So to make sure that I have the same amounts of each atom on both sides of my equation here, I'm going to add a coefficient of two in front of nitrogen oxide. So for this question, right, for this problem, we're going to take the mass of the reagent and we're going to use dimensional analysis to find the mass of the product that could be produced from the quantity of the re. Now, the challenging thing is that the masses of both reagents are actually provided. But recall that a question like this is asking us to find the theoretical yield of nitrogen oxide. The theoretical yield is determined from the limiting reagent. In other words, the theoretical yield is determined by the reagent that gets consumed entirely during the reaction. So, what we're going to do here is we're going to determine the mass produced from both reagents. And so the smaller amount of nitrogen oxide is the theoretical yield because that stems from the limiting reagent. But first, we have to recall the molar masses of our regions and our products, the molar mass of nitrogen is equal to 28.01 g per mole. The molar mass of oxygen gas is equal to 32.00 g per mole in the molar mass of nitrogen oxide is equal to 30.01 g per mole. So first, let's start with nitrogen. In this case, we're provided 2.75 g of N two, which we're going to first convert into moles of N two by using the molar mass. Now, when treating the conversion factor like a fraction, the idea is to always cancel out your starting units. So when using my molar mass grams of N two should go in the denominator. Whereas one mole of N two should go in the numerator. And the reason for this is to ensure that grams of N two cancel out. And now now that our units are moles of nitrogen, we can use the balanced chemical equation to relate the moles of N two to the moles of nitrogen oxide. So according to that two moles of nitrogen oxide are produced for every one mole of N two that's consumed. Once again, moles of N two cancel out. So now to convert our moles of nitrogen oxide to a mass, we're going to use the molar mass of nitrogen oxide. So that's 30.01 g of no in the numerator per one mole of no in the denominator once again, ensuring that moles of nitrogen oxide cancel out. So after evaluating this expression by multiplying all numerators and dividing by all denominators, our answer or the mass of nitrogen oxide expected from nitrogen is equal to five 0.8927 g of no. So now let's compare this to the mass of nitrogen oxide expected from the amount of oxygen provided. So to start off the second calculation, that's 8.79 g of 02 multiplied by the molar mass or rather divided by because one mole of 02 equals 32.00 g of 02 grams. Our place in the denominator to make sure that our starting units cancel out. So using the mole ratios from the balanced chemical equation, we can see that two moles of nitrogen oxide are produced for every one mole of 02 that's consumed. So now that moles of 02 have canceled out, we can use the molar mass of nitrogen oxide to express our answer in grams. So that's 30.01 g of no or every one mol. So after evaluating our expression, the mass of no expected from oxygen is equal to 16.4867 g of nitrogen oxide. So by comparing these masses to each other, we can see that N two is the limiting reagent because it produced a smaller amount of nitrogen oxide. Therefore, our theoretical yield is 5.8927 g, which rounded to three significant figures equals 5.89 g. And so if I scroll up, we can see that this matches option C in the multiple choice. So our answer is option C which is 5.89 g and there you have it. So if you stuck around until the end of this video, thank you so very much for watching. And I hope you found this helpful.