Question

What is the maximum amount in moles of P2O5 that can theoretically be made from 272 g of O2 and excess phosphorus? Express your answer to three significant figures and include the appropriate units.

Calculations involving a limiting reactant

Now consider a situation in which 25.0 g of P4 is added to 50.0 g of Cl2, and a chemical reaction occurs. To identify the limiting reactant, you will need to perform two separate calculations:

Calculate the number of moles of PCl5 that can be produced from 25.0 g of P4 (and excess Cl2).

Calculate the number of moles of PCl5 that can be produced from 50.0 g of Cl2 (and excess P4).

Then, compare the two values. The reactant that produces the smaller amount of product is the limiting reactant.

Answer 1

Sol.

(a) Reaction : P4 + 5O2 ----> 2P2O5  

As mass of O2 = 272 g

molar mass of O2 = 32 g / mol

So , moles of O2 = 272 / 32 = 8.5 mol  

From reaction , 5 moles of O2 give 2 moles of P2O5

So , 8.5 moles of O2 give

= 8.5 × 2 / 5 = 3.4 moles of P2O5

Therefore ,  3.4 mol is the answer

(b) Reaction : P4 + 10Cl2 ----> 4PCl5

mass of P4 = 25 g

molar mass of P4 = 123.90 g / mol

So , moles of P4 = 25 / 123.90 = 0.2018 mol

From reaction , 1 mole of P4 produces 4 moles of PCl5

So , 0.2018 moles of P4 produces = 4 × 0.2018 = 0.8072 moles of PCl5  

mass of Cl2 = 50 g

molar mass of Cl2 = 70.91 g / mol

moles of Cl2 = 50 / 70.91 = 0.7051 mol

From reaction , 10 moles of Cl2 produces 4 moles of PCl5

So ,0.7051 moles of Cl2 produces

= 0.7051 × 4 /10 = 0.2820 moles of PCl5

So , Cl2 reactant produces smaller amount ( moles ) of PCl5

Therefore ,   Cl2 is the limiting reactant and moles of PCl5 produced are 0.2820 mol