Question

What mass of p4010 is obtained from the reaction of 1.33g of p4 & 5.07 g of oxygen?

Answer 1

Supposing the missing units on "5.07" to be "grams":

(1.33 g P4) / (123.895048 g P4/mol) = 0.010735 mol P4

(5.07 g O2) / (31.99886 g O2/mol) = 0.15844 mol O2

P4 + 5 O2 → P4O10

0.010735 mole of P4 would react completely with 0.010735 x (5/1) = 0.053675 mole of O2, but there is more O2 present than that, so O2 is in excess and P4 is the limiting reactant.

(0.010735 mol P4) x (1 mol P4O10 / 1 mol P4) x (283.8890 g P4O10/mol) = 3.05 g P4O10.

Hope it helps ✌✌❤