Mass-Mass Problem (3?s)

Question

P4(s) 5O2(g) --> P4O10(g)

1. How many grams of phosphorus(IV) oxide (P4O10) is produced if you burn 50.0 grams of phosphorus with sufficient oxygen (O2)?

2. How many grams of oxygen would be needed in problem 1?

3. If 400 grams of phosphorus(IV) oxide (P4O10) is needed for another experiment, how much phosphorus would have to be burned?

Answer 1

The balanced equation is P4 + 4O2 = 2P2O4. Phosphorus weighs 31 grams.  Given 50 grams, divide by 31 = 1.6 moles.  From the equation, the ratio of P(lV) to P2O4 is 1:2.  If 1.6 represents your 1 then 3.2 moles represents your 2. The weight of P2O4 is:

P = 31 x 2 = 62 grams

O = 16 x 4 = 64 grams

Added together they total 126 grams per mole x 3.2 moles = 204.8 grams produced.  

Grams oxygen = 16 x 8 (from equation) = 128 grams

From the equation get the ratio again. P4 to P2O4 = 1:2.  If 400 grams represents your 2, then 200 grams equals your 1 or the amount of P4 needed.

Answer 2

Moles P4 = 50.0 g / 123.8952 g/mol = 0.404

The rario between P4 and P4O10 is 1 : 1

Moles P4O10 = 0.404

Mass P4O10 = 0.404 mol x 283.8892 g/mol = 114.7 g

The ratio between P4 and O2 is 1 : 5

Moles O2 needed = 0.404 x 5 =2.02

Mass O2 needed = 2.02 mol x 31.9988 g/mol = 64.6 g

Moles P4O10 = 400 g / 283.8892 = 1.41

Moles P4 needed = 1.41

Mass P4 = 1.41 mol x 123.8952 g/mol = 174.7 g