Question

P4+5O2→2P2O5

Consider a situation in which 112 g of P4 are exposed to 112 g of O2.

Part A

What is the maximum amount in moles of P2O5 that can theoretically be made from 112 g of P4 and excess oxygen?

Part B

What is the maximum amount in moles of P2O5 that can theoretically be made from 112 g of O2 and excess phosphorus?

Answer 1

A)

Molar mass of P4 = 123.88 g/mol

mass of P4 = 112 g

mol of P4 = (mass)/(molar mass)

= 112/123.88

= 0.9041 mol

From balanced chemical reaction, we see that

when 1 mol of P4 reacts, 2 mol of P2O5 is formed

mol of P2O5 formed = (2/1)* moles of P4

= (2/1)*0.9041

= 1.8082 mol

Answer: 1.81 mol

B)

Molar mass of O2 = 32 g/mol

mass of O2 = 112 g

mol of O2 = (mass)/(molar mass)

= 112/32

= 3.5 mol

From balanced chemical reaction, we see that

when 5 mol of O2 reacts, 2 mol of P2O5 is formed

mol of P2O5 formed = (2/5)* moles of O2

= (2/5)*3.5

= 1.4 mol

Answer: 1.40 mol