Question

consider the following reaction between phosphorus and oxygen gas:

P4(s) + 5O2(s) ---> P4010 (s)

A) how many moles of P4O10 could be prodiced from 16.0 g of P4 and 450.0mL of O2 at STP?

B) How many moles of which excess will react will remain at the end of the reaction?

Answer 1

P₄ + 5 O₂ = P₄O₁₀
Reaction type: synthesis

16 gm P4 = 16 / 123.89 = 0.12914 Mole

450.0mL of O2 at STP = 0.45 / 0.08206 x 273 = 0.02 Mole

Oxygen is limiting reagent

0.02 Mole of oxygen will react with (0.02/5) 0.004017 Mole P4 to give the same mole (0.004017 Mole) of P4O10

0.004017 Mole P4O10 = 0.004017 x 283.889 =   1.1355 gm of P4O10 will be produced

P4O10 will be excess = 0.12914 - 0.004017 = 0.125123 Mole x 283.889 = 35.52 gm P4O10 will be excess

0.125123 Mole of P4O10 will be excess