The industrial production of hydriodic acid takes place by treatment of iodine with hydrazine (N2H4): 2 I2 + N2H4 --> 4 HI + N2 (b) How many grams of HI are produced from the reaction of 115.7 g of N2H4 with excess iodine

Video transcript

Hey everyone. Our question here is asking us to calculate the g of product that form 123.59 g of oxygen gas completely reacts. Now, looking at our chemical reaction, we can see that it is completely balanced so we don't need to do any other work on that. And essentially, our question is asking us to go from grams of oxygen into moles of oxygen and relating that to moles of strontium oxide and then finally into grams of strontium oxide. Now that we have our pathway, let's go ahead and begin with our 23.59 g of oxygen gas. Now, in order to convert this into moles, we're going to need the molar mass of oxygen gas. So the molar mass of oxygen gas is going to be 16 times two, which is 32. In putting that into our dimensional analysis, we know that we have 32.0 g per one mole. And now looking at our reaction, we can see that per one mole of oxygen. We have two moles of strontium oxide and in order to convert this into grams of strontium oxide, we're going to need to calculate the molar mass of strontium oxide. And when we do that, we find that it's 103 .62 g per mole. Plugging that into our dimensional analysis. We know that per one mole of strontium oxide, we have 103.62 g of strontium oxide. And when we calculate this out, we end up with a value of .77 g of strontium oxide. And since we have four significant figures, We can round this out to 152. and this is going to be our final answer. So I hope this made sense and let us know if you have any questions.