Supposing we allow 272g of phosphorus to react with excess oxygen which forms tetraphosphorus decoxide P4O10 in 89.5% yield. In the second step reaction, a 96.8% yield of H3PO4 is obtain by reacting P4O10 with 6H2O. Work out the mass of H3PO4 obtained?

2980g

To work out the mass of H3PO4 obtained, we need to follow the steps of the reaction and calculate the yield at each step.

Step 1: Reaction between Phosphorus (P) and Oxygen (O) to form P4O10
The balanced equation for this reaction is:
P4 + 5O2 -> P4O10

According to the question, we have 272g of phosphorus. To calculate the mass of P4O10 formed, we need to determine the molar mass of P4O10:

Molar mass of P4O10 = (4 x atomic mass of P) + (10 x atomic mass of O)
= (4 x 31.0 g/mol) + (10 x 16.0 g/mol)
= 123.0 g/mol + 160.0 g/mol
= 283.0 g/mol

Now, we can calculate the theoretical yield of P4O10 using stoichiometry:

Theoretical yield of P4O10 = (mass of phosphorus / molar mass of P) * molar mass of P4O10
= (272g / 31.0 g/mol) * 283.0 g/mol
= 2328.39 g

However, the question states that the yield of P4O10 is only 89.5%. Hence, the actual yield of P4O10 is:

Actual yield of P4O10 = Theoretical yield * yield percentage
= 2328.39 g * 0.895
= 2084.69 g

Step 2: Reaction between P4O10 and H2O to form H3PO4
The balanced equation for this reaction is:
P4O10 + 6H2O -> 4H3PO4

According to the question, the yield of H3PO4 in this step is 96.8%. Now, we can calculate the mass of H3PO4 obtained:

Mass of H3PO4 = (Actual yield of P4O10 / molar mass of P4O10) * (molar mass of H3PO4 / yield percentage)
= (2084.69 g / 283.0 g/mol) * (98.0 g/mol / 0.968)
= 7.361 mol * 101.24 g/mol
= 744.54 g

Therefore, the mass of H3PO4 obtained is approximately 744.54 grams.