Assume that you have 1.39 mol of H2 and 3.44 mol of N2. How many ...

Question

Assume that you have 1.39 mol of H2 and 3.44 mol of N2. How many grams of ammonia (NH3) can you make, and how many grams of which reactant will be left over?

3 H2 + N2 --> NH3

Accepted Answer

Hello everyone. So in this video we're being asked first to determine the mass and grams of page three that can be produced. And the second question is to determine the mass and grams of the excess reactant. Let's first calculate the moles of H2 that's required to react with RP four because we have two different reagents here. So we're starting off with our 0.950 moles of P4. That's what we're given to us in the problem. Then we can just use the stark geometric rate multiple ratio too soft for the moles of H two required. So we have for every one mole of H or P four That we get six moles of our H2. We can see here that the moles of P4 will cancel Leaving us with 5.70 moles of H two. So we should see a little problem here. This is how much is needed to are required to react with RP four but were only given 3.95 moles of H two. So you can see here then that our age two is going to be our limiting reagent. So now with that in mind, let's go ahead and use our limiting reagent to calculate the mass of ph three can be produced. Alright, so let's first also realize that the Mueller mass of our ph three is equal to 33.998 g promote. Alright, we can now start our calculations. So we have again using our limiting region of 3.95 moles of our H two. We can use these third geometric multiple ratio to convert our moles of H two into the mold of our product. So again we have for every four moles of our ph three we get six moles of our H. Two. You can see here the moles of H two council. So now using the molar mass of our product, we can change our moles into grams. So for every one mole of p three We have 33.98 g of Ph three. You can see here how the moles of PhD people cancel. So putting all these numbers into my calculator, I'll go ahead and get the value of 89. g of ph three. So we have successfully answered the first part of question. Now moving on to the second. So let's first calculate for the initial mass of P four which want to convert the given moles into our grams units. So of course when you're going to need the molar mass, the molar mass of P four is equal to 1 to 3.896. Of course units being grams per mole. Alright, so the mass of P four initially. So let's first start with the given moles of 0.95 moles Of P four. We're going to go ahead and use the molar mass to give us the g of the product. So we can see here we do that. We can say for every one mole of RP four we get 1 to 3.896 g of P four. Again, of course the multi P four will counsel giving us the market value. Let's see here putting this in the calculator. We get 117.7012 g of P four. Alright now calculate for the mass of P four that has actually reacted are used basically. So we're starting with the lemony reagent which is 3.95 moles of H two. Again using the stock geometric multiple ratio. So for every six moles of H two we get one mole of R. P four. So the moles of H two will cancel. Then going from moles of P four and two g of P four would use the molar mass. So again, mole of P four will give us 123.896 g of P four. Again, the moles of P four will cancel Giving us the answer of 81.5649 g of P four. Now finally calculating for the massive p. four that remains which is our main question. So that's just the initial minus was used to give us the remaining. So again we have 117.7012 g. That's how much originally had. Then we used 81.5649 g giving us the remainder of 36.1 g of P four. So that successfully answers our second question here and the problem. Thank you all so much for watching.

Assume that you have 1.39 mol of H2 and 3.44 mol of N2. How many grams of ammonia (NH3) can you make, and how many grams of which reactant will be left over? 3 H2 + N2 --> NH3

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