SS2 Chemistry Lesson Note on Calculations on Mass- Volume

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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 P_{4   =} 31 × 4 = 124gmol^-1

O₂ =16 × 2 = 32 g.mol^{– 1}

P₄ O_{6 =}  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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