Explanation:
The Limiting Reactant is that reactant which when consumed in a reaction stops the reaction. The other reactants will be in excess and typically considered non-reactive.
To identify the limiting reactant ...
- write and balance the reaction of interest. Express it in standard form. That is, standard form of a reaction is when the coefficients of the balanced equation are in their lowest whole number values. Also, remember that the standard equation is 'assumed' to be at STP conditions (0°C & 1atm).
- convert all given reactant values to moles
- divide each reactant mole value by the related coefficient of the the balanced standard equation. The smaller value is the limiting reactant. The remaining reactants will be in excess.
Your Problem:
Given: 3Ba + N₂ => Ba₃N₂
22.6g 4.2g ?
moles Ba => 22.6g/137.34g/mol = 0.165 mole Ba
moles N₂ => 4.2g/14.007g/mol= 0.150 mole N₂
Part A: Determining the Limited Reactant
Divide each mole value by respective coefficient ... smallest value is Limiting Reactant.
Barium => 0.165/3 = 0.055 <=> (Limiting Reactant)
Nitrogen => 0.15/1 = 0.15
Barium is the smaller result and is therefore the limiting reactant. This works for ALL limiting reactant type problems. However, be sure to use the mole values calculated first (Ba = 0.165mol & N₂ = 0.150mol) when doing ratio calculations.
Part B: Max (theoretical) amount of Ba₃N₂ produced:
Note: The product yield amounts are based upon the given 'moles' of limiting reactant, NOT the results of the 'divide by respective coefficient' step used to ID the limiting reactant.
3Ba + N₂ => Ba₃N₂ (3:1 rxn ratio for Ba:Ba₃N₂)
moles 0.165mole 0.150mole 1/3(0.165)mole = 0.055mole Ba₃N₂
= 0.055mol(440g/mol) Ba₃N₂
= 24.2 grams Ba₃N₂ (as based
upon Barium as Limiting Reactant)
Part C: Excess N₂ remaining after reaction stops:
From balanced standard reaction, the reaction ratio for Ba:N₂ is 3moles:1mole. That is, for the moles of Ba consumed, 1/3(moles of Ba) = moles of N₂ used.
moles of N₂ used = 1/3(0.165)mole = 0.055mole N₂ used
∴ the amount of N₂ remaining in excess = 0.150mole (given) - 0.055mole (used) = 0.095mole N₂ remaining in excess.
mass N₂ remaining = 0.095mole x 28g/mole = 2.66 grams N₂ remaining in excess.
