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Report (any type)/Brief report
22/09/2021
Once you have identified the limiting reactant, you calculate how much of the other reactant it must have reacted with and subtract from the original amount.
A Sandwich-Making Analogy
This video from Noel Pauller uses the analogy of making sandwiches.
The general problem
Given the chemical equation and the masses of reactants, determine the mass of excess reactant and the mass of the limiting reactant required to use up the excess.
A specific problem
A 2.00 g sample of ammonia reacts with 4.00 g of oxygen according to the equation
##”4NH”_3 + “5O”_2 → “4NO” + “6H”_2″O”##.
How much excess reactant remains after the reaction has stopped? How much more of the limiting reactant would you need to use up the excess?
Strategy
Solution
1. Balanced equation
##”4NH”_3 + “5O”_2 → “4NO” + “6H”_2″O”##
2. Moles of ##”NO”## from ##”NH”_3##
Convert grams of ##”NH”_3## to moles of ##”NH”_3##, and then use the molar ratio from the equation to get moles of ##”NO”##.
The molar mass of ##”NH”_3## is 17.03 g/mol.
##”Moles of NH”_3 = 2.00 cancel(“g NH”_3) × (“1 mol NH”_3)/(17.03 cancel(“g NH”_3)) = “0.1174 mol NH”_3##
##0.1174 cancel(“mol NH₃”) × “4 mol NO”/(4 cancel(“mol NH₃”)) = “0.1174 mol NO”##
3. Moles of ##”NO”## from ##”O”_2##
The molar mass of ##”O”_2## is 32.00 g/mol.
##”Moles of O”_2 = 4.00 cancel(“g O”_2) × (“1 mol O”_2)/(32.00 cancel(“g O”_2)) = “0.1250 mol O”_2##
We know from the balanced equation that the molar ratio is ##”4 mol NO ≡ 5 mol O”_2##, so we create a conversion factor with “##”mol O”_2##” on the bottom to make the units cancel.
##0.1250 cancel(“mol O”_2) × “4 mol NO”/(5 cancel(“mol O”_2)) = “0.1000 mol NO”##
4. Identify limiting and excess reactants
##”O”_2## is the limiting reactant, since it gives the smaller amount of ##”NO”##.
##”NH”_3## is the only other reactant, so it is the excess reactant.
5. Calculate the mass of excess reactant used up.
Use the molar ratio from the equation to convert moles of ##”O”_2## (from Step 3) to moles of ##”NH”_3##, and then convert moles of ##”NH”_3## to grams of ##”NH”_3##.
##0.1250 cancel(“mol O”_2)× (“4 mol NH”_3)/(5 cancel(“mol O”_2)) = “0.1000 mol NH”_3##
##0.1000 cancel(“mol NH”_3) × (“17.03 g NH”_3)/(1 cancel(“mol NH”_3)) = “1.703 g NH”_3##
6. Calculate the mass of unused excess reactant.
We started with 2.00 g of ##”NH”_3## and used up 1.703 g, so
##”Mass of excess NH”_3 = “2.00 g – 1.703 g” = “0.30 g”##
7. Calculate the mass of limiting reactant needed to react with the leftover excess reactant.
##”Moles of NH”_3 = 0.30 cancel(“g NH”_3) × (“1 mol NH”_3)/(17.03 cancel(“g NH”_3)) = “0.0176 mol NH”_3##
##”Moles of O”_2 = 0.0176 cancel(“mol NH”_3) × (5 cancel(“mol O”_2))/(4 cancel(“mol NH”_3)) = “0.0220 mol O”_2##
##”Mass of O”_2 = 0.0220 cancel(“mol O”_2) × (“32.00 g O”_2)/(1 cancel(“mol O”_2)) = “0.70 g O”_2##
Answer
It takes 0.70 g of ##”O”_2## to react with the 0.30 g of excess ##”NH”_3##.
Here is another example…
I want to thank your team (Jeff) for finding a new writer when you were unable to reach my existing writer to revise my paper. I appreciate your professionalism and efforts. Thanks
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