Question

Phosphorus burns in a limited amount of oxygen to produce a compound that is 56.4% P and 43.6% O by mass. How much phosphorus (as P4) and oxygen (as O2) remain when 10.00 g P4 reacts with 5.00 g O2? Please provide a brief explanation in sentences as well please

Answer 1

Assume that the amount of compound is 100 g

Then there is 56.4 g P,

43.6 g O

now calculate the number o f moles:

Number of moles of P= 56.4 g/30.97 g/ mol

= 1.82 mol P

Number of moles of O= 43.6 g / 15.99 g/ mol

= 2.73 mole O

Now determine the mole ratio:

P= 1.82 /1.82 = 1

O= 2.73/1.82 =1.5

Now change digits into whole number by multiplying 1,2 and so on.

First multiply by 2,

P= 1*2=2

O=1.5*2=3

Thus the empirical formula is P2O3, of product

Thus the balance reaction when Phosphorus burns in a limited amount of oxygen is as follows:

P₄ + 3O₂ = 2P₂O₃

Now calculate the number of moles of P and O as follows:

Number of moles of P = 10.0 g/ 123.90 g/mol

= 0.081 moles P4

And oxygen ; 5.00 g / 31.99880 g/ mole

= 0.156 moles

In this reaction and amount O2 is a limiting agent

The limiting agent has due to following properties:

1. It completely reacted in the reaction.
2. It determines the amount of the product in mole.

Thus after competition reaction there is no O2.

Now calculate the amount of P4 used:

0.156 moles O2 * 1 mole P₄ / 3 mole O₂

= 0.052 moles P4

Then calculate the moles of P4 which is remain in the reaction;

Total moles of P4 – used mole of P4

= 0.081 moles P4 -0.052 moles P4

= 0.029 moles P4

Amount in g = number of mole s* molar mass

= 0.029 moles P4 *123.90 g/mol

=3.5931 g P4

= 3.6 g P4