If 2.00 moles of nitrogen and 5.50 moles of hydrogen are placed i...

Question

If 2.00 moles of nitrogen and 5.50 moles of hydrogen are placed in a reaction vessel and react to form ammonia, what is the theoretical yield of ammonia (NH3)? (LO 3.8) 
N2(g) + 3 H2(g) --> 2 NH3(g)
(a) 31.2 g 
(b) 62.3 g
(c) 93.7 g 
(d) 34.1

Accepted Answer

Hi everyone for this problem will be calculating the theoretical yield of a gcl if a g n 03 and K C L are allowed to react. So they've provided us with our chemical equation, it's up here, let's just make sure it's balanced real quick. We have won a G one, a G one N 031 N 03. We have one K one K and one cl one cl. So we're all balanced. Let's start by finding how much a gcl can be formed from each of our reactant. So we have 0.78 moles Of AGN 03. And all we need to do is multiply it by the multiple ratio between a g n 03 and a gcl. So for every one mole a g n 03, we have one mole a gcl. So one mole a gcl for one more AGN 03. So this still leaves us with 0.78 moles of a gcl that can be formed. So we're going to do the same thing With our case cl so we have 0.39 moles of k C L. And we're going to multiply it by the multiple ratio. So for every one mole K C L, we have one mole a gcl. So one mole a gcl, one mol K c L. So that's still going to give us 0.39 moles of a gcl and that's what can be formed. So now all we have to do is convert This .39 moles because KCL is going to be our limiting reactant since it gave us the least amount of moles. So we're going to convert that 0.39 malls of a gcl and we're going to convert it to grams. So we need to multiply it by its molar mass conversions. So we're given this actually, it's going to be 1 43.3. So 1 43.3 g of a G C L for one mole of a G C L. When you solve it out, you'll find that we have 55.89 g of a gcl that's going to be formed here. When we take a look at our options, we'll see that B is the correct answer. I hope this video helped you guys out and I'll see you in the next one.

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