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MORE CALCULATIONS ON MASS- VOLUME
Phosphorous burns in the limited supply of air to form phosphorous(iii) oxide, P4O6 and in unlimited supply of air to form phosphorous(v) oxide P4O10.If 3.3g of phosphorus(III) oxide is formed
(a) what mass oxygen from air is used?
(b) How many molecules of oxygen are contained in this mass?
(c) How many molecules of phosphorous are converted to the oxide?
(d) How many molecules of oxygen are needed to convert this oxide into phosphorous v oxide?
(e) What volume would the oxygen in (d) above occupy at s.t.p.?
Solution:
Molar mass of P4 = 31 × 4 = 124gmol-1
O2 =16 × 2 = 32 g.mol– 1
P4 O6 = gmol–
- 220g of P4 O6 used (32 ×3) 96g of oxygen
3.3 of P4O6 requires g of oxygen
= 1.44g
(b) 1.44g of oxygen = mole
1 mole contain 6.02 × molecules
Number of moles in 1.44g of oxygen =
= 2.71 molecules
(c) Mass of phosphorous in 3.3g of P4O6 = (3.3 – 1.44)g
= 1.86g
1.86 g of phosphorous is mole
The number of molecules in (c) above i.e. 1.86g of phosphorus
= = 9.03 molecules
- P4O6 +202 P4O10
220 of P4O6 require 2 moles of oxygen
3.3g of P4O6 will require
Number of molecules of oxygen contained in this molecule
= 1.81 molecules
- 02 molecules occupy 22.4dm3 at s.t.p
1.81 molecule occupy 22.4dm3
= 0.067 dm3 or 6.73.49 cm3
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