Question

What volume of oxygen, in Liters, is needed to react with 18.65 g of P4?

P4 + 5O2 → P4O10

Answer 1

Answer: 16.8L (at STP)

Explanation:

Mole P4O10 = 4*31+10*16 = 284

P4 = 124

moles P4O10 = 18.65/124 = 0.15

5 moles O2 per mole P4O10, so 5*0.15 moles O2 required = 0.75 moles

at STP 1 mole gas = 22.4 L

vol O2 = 0.75*22.4 = 16.8 L