Question

How many grams of P205 will be produced when 17.0 g of PH3 is mixed with 16.0 g of O2 in the following reaction? 2PH3 + 402 → P205 + 3H20 O 17.08 O 56.4 g O 35.5 g O 71.0 g 54.7 g

Answer 1

2PH3 + 4O2 ----> P2O5 + 3H2O

Moles of PH3 = mass/molar mass = 17/34 = 0.5

Moles of O2 = mass/molar mass = 16/32 = 0.5

From reaction;

2 moles PH3 requires 4 moles O2

So, 0.5 mole PH3 will require = 4*0.5/2 = 1 mole O2

However available moles of O2 = 0.5 which is less than the required one i.e. 1 mole

Hence O2 is limiting reagent and it will drive the yield of P2O5

Again from reaction;

4 mole O2 produces 1 mole P2O5

So, 0.5 mole O2 will produce = 1*0.5/4 = 0.125 moles of P2O5

Mass of P2O5 = moles * molar mass = 0.125 * 283.886 = 35.49 g OR 35.50 grams ....Answer