Limiting Reagent

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Katherine Wu 1H
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Joined: Fri Aug 30, 2019 12:15 am
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Limiting Reagent

Postby Katherine Wu 1H » Fri Sep 27, 2019 7:21 pm

For the following equation, determine the limiting reagent if 21.4 g NH3 is reacted with 42.5 g of O2.
4NH3(g) + 5O2(g) ---> 4NO(g) + 6H2O(g)
After dividing both of their weights by their molar masses, I end up with NH3=1.257 mol and O2=1.328 mol. I'm confused on what to do afterward.

Harry Zhang 1B
Posts: 101
Joined: Sat Sep 14, 2019 12:16 am

Re: Limiting Reagent

Postby Harry Zhang 1B » Fri Sep 27, 2019 7:34 pm

Your previous steps are correct, so I will just answer what to do next. After getting the number of moles of each reactant, you should use stoichiometry to determine the number of moles of either one of the product produced. Let's say we choose NO(g) as the product that we will be using to compare the two reactants. Using stoichiometry, the theoretical(maximum) number of NO that can be produced is 1.257mol NH3 x 4 mol NO/4 mol NH3=1.257 mol NO(g). For O2, the theoretical(maximum) number of NO that can be produced is 1.328 mol O2 x 4 mol NO/5 mol O2=1.062 mol NO(g). From these two equations, we can see that the number of NO produced from O2 is smaller than that produced from NH3, which makes O2 the limiting reactant since the maximum number of product that can be produced, in this case, NO, is limited by O2 to be 1.062 mol. After producing this much product, the oxygen gas in the chemical reaction is depleted and no more product can be formed despite there are still some NH3 in excess.

Hope this helps!

WesleyWu_1C
Posts: 117
Joined: Thu Jul 25, 2019 12:16 am

Re: Limiting Reagent

Postby WesleyWu_1C » Sat Sep 28, 2019 3:31 am

There is another way to determine the limiting reactant from your work. Lets imagine that all 1.257 mole of NH3 were used up. If all 1.257 moles of NH3 were used up, how much moles of O2 were also needed for that reaction? We can answer this question by using 1.257 mol of NH3 and multiplying by the mole ratio of O2 to NH3 (5 moles to 4 moles), which is found in the balanced chemical equation. This gets 1.570 mole of O2. So to review this number, 1.570 mol of O2 means that if all 1.257 mol of NH3 were used, there would need to be at least 1.570 mol of O2 in the reaction. However, if we compare the 1.570 mol of O2 to the actual number of moles of O2 (1.328 mol), we see that we have less moles of O2 required to use up all of the NH3. So we do not have enough O2 to use up all the NH3, which in other words means that O2 is the limiting reactant and that if this reaction would occur we would have left over NH3.


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