Calculate how many moles of NO₂ form when each quantity of reactant completely reacts. 2 N₂O₅( g) → 4 NO₂(g) + O₂(g) a. 2.5 mol N₂O₅

Video transcript

Hey everyone. Our question here asked if 5.7 moles of di nitrogen Penta oxide react, how many moles of nitrogen dioxide form, Looking at our reaction, we can see that this is unbalanced. So let's go ahead and balance this out first before answering our question. So in our react inside we have two of nitrogen and five of oxygen and in our product side we have one of nitrogen And four of oxygen. Now, in order to balance this out, we're going to need to change our oxygen in our reacting side to be an even number. So we can do this by multiplying our react inside by two. Now this will change our nitrogen into four And our oxygen's into 10. Since we did this, we're going to have to balance out our product side as well. So by multiplying nitrogen dioxide by four, We end up with four of nitrogen and a total of 10 oxygen's which gives us our balanced chemical equation. Now we can go ahead and move on with our question. Starting off with 5.7 mol of di nitrogen Penta oxide, we can see that in our balance chemical reaction We have two moles of di nitrogen Penta oxide and this is equivalent to formals of nitrogen dioxide. Now, when we calculate this out, We end up with a total of 22.8 mole of nitrogen dioxide, Which we can round up to 23 and this is going to be our final answer. So I hope this made sense and let us know if you have any questions