Nitrogen and hydrogen combine at a high temperature, in the prese...

Question

Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia.

Accepted Answer

Hi everyone. So for this problem, we need to calculate the moles of ammonia that's going to be produced, the moles of H two that will remain the molds of end to that will remain. And we also need to find the limiting reactant were already given a balanced chemical equation. So we can jump right into our first part of finding the amount of ammonia produced. So to do this, we're going to have to calculate how much can be produced from each reactant. So we'll start with our 0.283 moles of end to and we need to multiply that by the multiple ratio between N two and ammonia. So we have one mole of N two for two moles of NH three. So we're going to put two moles of NH three for one mole of N two And then we can cancel our units. And we'll find that we are going to be left with 0.566 moles of ammonia. And that's what we can produce from our amount of end to next. We'll do the same thing for our H two. So we're starting with 0.911 moles of H two. And our multiple ratio is going to be three moles of H two for two moles of ammonia. So two moles NH three for three moles of H two. So we can cancel out our units again. And we're going to find that we're going to be left with 0.607 moles of ammonia and that's what we're going to be creating from our amount of hydrogen. So from this, we actually get a lot of information. So because .566 is going to be our smaller amount. That is what we're going to be able to make total because since we have a limited amount of end to, that's the reaction that produced it. That's going to be the most we can make. So we have our moles of ammonia That's going to be 0. malls. And this also tells us are limiting reactant because end to produce the least amount of moles and two is going to be our limiting reactant. And it also tells us the moles of end to that are going to be left over. Because if it's our limiting reactant, we're going to use all of it up in this equation. So there are going to be zero moles of end two left over. This leaves us with just the moles of H2 left over that we have to calculate. So let's go ahead and do that. So first things first to find this, we first need to convert our 0.566 moles of ammonia back two moles of H two. So we take 0.566 moles of ammonia and we're going to multiply it by our multiple ratio. So three moles of H two for two moles of ammonia. So we're going to put three moles H two for two moles of ammonia And then cancel our units. And when you multiply this out, you'll find that we have 0. malls of H two and that's what's going to be used. That's what we're using to form our ammonia. Now, all that's left to do is we're going to subtract the amount that we used from the amount we started with. So we started with 0.911 moles of H two. And we're going to subtract 0.849 moles of H two that we used. And when you solve that out, you're going to be left with 0.062 moles Of H two. That's going to be left. So we can go ahead and put this back in with our answers. So the moles of Ammonia or I'm sorry, moles of H2 left is going to be 0.06 two moles each two. So this box over here on the right, this is going to be our answer to this problem. I hope this video helped you out and I'll see you in the next one

Nitrogen and hydrogen combine at a high temperature, in the presence of a catalyst, to produce ammonia.

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