Limiting Reactant  [ENDORSED]

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ali_boutrosdisc3H
Posts: 18
Joined: Wed Sep 21, 2016 2:59 pm

Limiting Reactant

Postby ali_boutrosdisc3H » Sun Oct 09, 2016 2:10 pm

In a limiting reactant problem, how do you know which mol ratios you're supposed to compare?

For example #9 on the self test in the workbook: for the following equation, determine the limiting reagent is 22.4gNH3 is reacted with 40.5g of CO2

NH3(g) + O2 (g) ---> NO(g) + H20(g)

Casey Monahan 2C
Posts: 18
Joined: Wed Sep 21, 2016 2:57 pm

Re: Limiting Reactant  [ENDORSED]

Postby Casey Monahan 2C » Sun Oct 09, 2016 7:20 pm

You find the number of moles of NO that NH3 will produce, then find the number of moles of NO that CO2 will produce, and whichever amount produces the least amount of NO will be the limiting reactant. Basically, you take the two reactants and find out which one produces the least amount of product.

Kevin Tam 1J
Posts: 23
Joined: Wed Sep 21, 2016 2:59 pm

Re: Limiting Reactant

Postby Kevin Tam 1J » Mon Oct 10, 2016 6:08 pm

Another way of doing this problem is to take one of the reactants and calculate the required amount of the second reactant using a mole ratio.

(22.4g NH3/17.03g mol^-1 NH3) = 1.32 mol NH3
40.5g O2/32.00g mol^-1 O2) = 1.27 mol O2

Let's take 1.32 mol NH3 and multiply it by the mole ratio, 5 mol O2/4 mol NH3.

1.32 mol NH3 x (5 mol O2/4 mol NH3) = 1.65 mol O2 required. Since only 1.27 mol O2 were provided and the amount required to react was 1.65 mol O2, O2 is the limiting reactant.

Note: Casey Monahan's way of approaching this problem is different but correct.


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