Consider a situation in which 211 g

g

of P4

P

4

are exposed to 240 g

g

of O2

O

2

. What is the maximum amount in moles of P2O5

P

2

O

5

that can theoretically be made from 211 g

g

of P4

P

4

and excess oxygen?

Express your answer to three significant figures and include the appropriate units.

3.00 mol

Explanation:

Given dа ta:

Mass of P₄ = 211 g

Mass of oxygen = 240 g

Moles of P₂O₅ = ?

Solution:

Chemical equation:

P₄ + 5O₂ → 2P₂O₅

Number of moles of P₄:

Number of moles = mass / molar mass

Number of moles = 211 g / 123.88 g/mol

Number of moles = 1.7 mol

Number of moles of O₂:

Number of moles = mass / molar mass

Number of moles = 240 g / 32g/mol

Number of moles = 7.5 mol

Now we will compare the moles of product with reactant.

O₂ : P₂O₅

5 : 2

7.5 : 2/5*7.5 = 3.00

P₄ : P₂O₅

1 : 2

1.7 : 2*1.7 = 3.4 mol

Oxygen is limiting reactant so the number of moles of P₂O₅ are 3.00 mol.

Mass of P₂O₅:

Mass = number of moles * molar mass

Mass = 3 mol * 283.9 g/mol

Mass = 852 g