Question

How many grams of air are required to complete the combustion of 186 g of phosphorus (M= 31 g/mol) to diphosphorus pentoxide, assuming the air to be 23% oxygen by mass?

4P(s) + 5O2(g) — 2P2Os(s)

Answer 1

1043.5 g of air to the nearest tenth.

Step-by-step explanation:

4P + 5O2 ---> 2P2O5

4*31 g P reacts with 5*32 g oxygen

124 g .. .. . .... . 160g oxygen

186 g .. .. .. .. .. (160*186)/124 g oxygen

- so the mass of air required = [(160*186)/124] / 0.23

= 240 / 0.23

= 1043.5 g of air.