find the limiting reagent for each set of reactants 4NH3(g) +3O2(g) = 2N2(g) + 6H2O(l) if you have 7.00 mol NH3 + 5.00 mol O2
Use the coefficients to convert mols of each to mols of any product; e.g. N2.
7 mols NH3 x (2 mols N2/4 mols NH3) = 7/2 = 3.5 mols N2.
5 mol O2 x (2 mols N2/3 mols O2) = 10/3 = 3.33 mol N2.
So the smallest amount of N2 will form which makes O2 the limiting reagent. That
s the long way.
Short way. 7 mol NH3 will require 7*3/4 mol O2 or 21/4 = 5.25 but you don't have tht much O2 so O2 must be the LR.
To find the limiting reagent for the given reaction, you need to compare the stoichiometric ratios between the reactants. The reactant that is completely consumed (runs out first) is the limiting reagent.
Let's determine the limiting reagent by comparing the moles of NH3 and O2 present:
4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(l)
For 4NH3, you have 7.00 mol NH3.
For 3O2, you have 5.00 mol O2.
Now, we will calculate the moles of N2 produced from each reactant:
From 7.00 mol NH3, you can produce 7.00/4 = 1.75 mol N2.
From 5.00 mol O2, you can produce 5.00/3 = 1.67 mol N2.
Since you can produce a lesser amount of N2 from the O2 reactant, O2 is the limiting reagent. Therefore, O2 would be the limiting reagent in this reaction.
To find the limiting reagent, you need to compare the number of moles of each reactant to the stoichiometric coefficients in the balanced equation. The reactant that yields the smallest amount of product is the limiting reagent.
Let's start by calculating the number of moles of N2 and H2O that can be produced from 7.00 mol NH3:
From the balanced equation, we can see that 4 moles of NH3 react to produce 2 moles of N2. Therefore, using the given 7.00 mol of NH3, we have:
(7.00 mol NH3) * (2 mol N2 / 4 mol NH3) = 3.50 mol N2
Similarly, we can calculate the moles of H2O produced from 7.00 mol NH3:
(7.00 mol NH3) * (6 mol H2O / 4 mol NH3) = 10.50 mol H2O
Now, let's calculate the number of moles of N2 and H2O that can be produced from 5.00 mol O2:
From the balanced equation, we can see that 3 moles of O2 react to produce 2 moles of N2. Therefore, using the given 5.00 mol of O2, we have:
(5.00 mol O2) * (2 mol N2 / 3 mol O2) = 3.33 mol N2
For H2O, we can directly use the balanced equation as there is no O2 involved:
(5.00 mol O2) * (6 mol H2O / 3 mol O2) = 10.00 mol H2O
To determine the limiting reagent, we compare the moles of each product obtained from NH3 and O2. In this case, we can see that the mole ratio of N2 derived from NH3 is 3.50 mol, while the mole ratio of N2 derived from O2 is 3.33 mol. Since 3.33 mol is smaller than 3.50 mol, O2 is the limiting reagent for the reaction.
Similarly, comparing the mole ratio of H2O derived from NH3 (10.50 mol) and from O2 (10.00 mol), we find that NH3 is the limiting reagent for the production of H2O.
Therefore, in this reaction, O2 is the limiting reagent for the production of N2, and NH3 is the limiting reagent for the production of H2O.