7.55 grams of P 4 and 7.55 grams of O 2 react according to the following reactio

Question

7.55 grams of P4 and 7.55 grams of O2 react according to the following reaction:

P4 + O2--> P4O6

If enough oxygen is available, then the P4O6 reacts further:

P4O6 + O2 --> P4O10

a. a. Find the limiting reagent in the formation of P4O10.

b. b. What mass of P4O10 is produced?

c. c. What mass of excess reactant remains?

Make sure to show ALL work, use unit conversion, show units, and walk me through the problem to arrive at the correct answer. Make sure the work is correct as well, please. Thanks so much!

Explanation / Answer

First, balance both rxns:

P4 + 3O2 = P4O6

P4O6 + 2O2 = P4O10

Next, combine both rxns to form an overall rxn:

P4 + 3O2 + P4O6 + 2O2 = P4O6 + P4O10

P4 + 5O2 + P4O6 = P4O6 + P4O10 ... P4O6 will cancel

P4 + 5O2 = P4O10

a)

First, solve for the moles of both P4 and O2:

Moles of P4 = 7.55 g x 1 mol / 123.90 g = 0.0609 moles

Moles of O2 = 7.55 g x 1 mol / 32.00 g = 0.236 moles

Next, solve for the moles of P4O10 from the moles of both P4 and O2 to figure out the limiting reagent:

Moles of P4O10 from moles of P4 = 0.0609 moles of P4 x 1 mole of P4O10 / 1 mole of P4 = 0.0609 moles

Moles of P4O10 from the moles of O2 = 0.236 moles of O2 x 1 mole of P4O10 / 5 moles of O2 = 0.0472 moles

Since O2 produces the fewer moles of P4O10, it is the limiting reagent

b)

Theoretical mass of P4O10 = 0.0472 moles x 283.89 g / mol = 13.4 g

c)

Amount of excess reactant (P4) in g that reacted with the limiting reactant (O2):

7.55 g of O2 x 1 mol of O2 / 32.00 g of O2 x 1 mole of P4 / 5 moles of O2 x 123.90 g of P4 / 1 mole of P4 = 5.85 g

Amount of excess reactant remaining:

7.55 g of P4 (original sample) - 5.85 g (reacted) = 1.70 g

Hope this helps and good luck! :)

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