Answer:
34.8 g
Explanation:
Answer:
We have the masses of two reactants, so this is a limiting reactant problem.
We will need a balanced equation with masses, moles, and molar masses of the compounds involved.
1. Gather all the information in one place with molar masses above the formulas and masses below them.
Mᵣ: 123.90 70.91 208.24
P₄ + 20Cl₂ ⟶ 4PCl₅
Mass/g: 46.0 32.0
2. Calculate the moles of each reactant
3. Calculate the moles of PCl₅ we can obtain from each reactant
From P₄:
The molar ratio is 4 mol PCl₅:4 mol P₄
From Cl₂:
The molar ratio is 4 mol PCl₅:20 mol Cl₂
4. Identify the limiting and excess reactants
The limiting reactant is chlorine, because it gives the smaller amount of PCl₅.
The excess reactant is phosphorus.
5. Mass of excess reactant
(a) Moles of P₄ reacted
The molar ratio is 1 mol P₄:20 mol Cl₂
(b) Mass of P₄ reacted
(c) Mass of P₄ remaining
Mass remaining = original mass – mass reacted = (46.0 - 11.18) g = 34.8 g P₄