Calculate: (a) The amount, in mol, of white phosphorus used, (b) the limiting reagent, (c) the excess amount, in mol, of the other reagent.

P₄ (s) + 3OH^- (aq) + 3H₂O(l) ---> PH₃ (g) + 3H₂PO₂^-(aq)
Provided for answering (a): Mass of white phosphorus: 2.478gProvided for answering (b): Volume of aqueous sodium hydroxide: 100cm³Concentration of aqueous sodium hydroxide: 5.00 moldm^-3

<--- ANSWER --->
P₄ (s) + 3OH^- (aq) + 3H₂O(l) ---> PH₃ (g) + 3H₂PO₂^-(aq)
(a) N= M/Mr ∴ 2.478/(30.97*4) = 0.02 mol(b) N= c * v ∴ 5 * (100/1000) = 0.5 mol∴ P₄ is limiting. (c) The other reagent is H₂O. So first, you use the mol (N) of the limiting reagent. So, we do: 0.02 mol * 3 (because there are 3 moles of H₂O.) = 0.06 mol Then, using the mol (N) of the excess reagent, we do: 0.5 - 0.06 = 0.440 mol.