Etermine the percent yield of the following reaction when 2.80 g of p reacts with excess oxygen. the actual yield of this reaction is determined to by 3.89 g of p2o5.
4 p+5o2→2p2o5

Yield = 66.7%

Explanation:

4P + 5O₂ → 2P₂O₅

P: 31 g/mol

O₂: 32 g/mol

P₂O₅: 142 g/mol

31 g P 1 mol

2.80 g P __     x

  x = 0.09 mol P

142 g P₂O₅ 1 mol

3.89 g P₂O₅ ___     y

     y = 0.03 mol P₂O₅

In a 100% yield, 4 mol of P produces 2 mol of P₂O₅. 0.09 of P should, therefore, produce 0.045 mol of P₂O₅.

Yield = (0.03 mol / 0.045 mol) x 100

Yield = 66.7%

the dependant variable is spilled water

the independent is the heat

50.23 i believe is correct but don't quote me on that.