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. Â
Your Problem:
Given:     3Ba  +  N₂  => Ba₃N₂
        22.6g   4.2g     ?
moles Ba => 22.6g/137.34g/mol = 0.165 mole Ba
moles Nâ‚‚ => Â Â 4.2g/14.007g/mol= 0.150 mole Nâ‚‚
Part A: Determining the Limited Reactant
Divide each mole value by respective coefficient ... smallest value is Limiting  Reactant.
Barium => 0.165/3 = 0.055 Â <=> (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 & Nâ‚‚ = 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. Â Â
          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 Nâ‚‚ remaining after reaction stops:
From balanced standard reaction, the reaction ratio for Ba:Nâ‚‚ is 3moles:1mole. That is, for the moles of Ba consumed, 1/3(moles of Ba) = Â moles of Nâ‚‚ used.
moles of Nâ‚‚ used = 1/3(0.165)mole = 0.055mole Nâ‚‚ used Â
∴ the amount of N₂ remaining in excess = 0.150mole (given) - 0.055mole (used) = 0.095mole N₂ remaining in excess.
mass Nâ‚‚ remaining = 0.095mole x 28g/mole = 2.66 grams Nâ‚‚ remaining in excess.
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