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The reaction between potassium chlorate and red phosphorus takes place when you strike a match on a matchbox. If you were to react 37.1 g of potassium chlorate (KClO_3) with excess red phosphorus, what mass of tetraphosphorus decaoxide (P_4O_{10}) could be produced?

Answer :

25.55 grams of tetraphosphorus decaoxide could be produced by the reaction.

Explanation:

First the balanced chemical reaction of the production of tetraphosphorus decaoxide is to be known.

The chemical equation is

10 KClO3 + 3P4⇒ 3 P4010 + 10 KCl

The number of moles of KCLO3 will be calculated by the formula:

number of moles = mass of the compound given ÷ atomic mass of the compound

n = 37.1 ÷ 122.55 ( atomic mass of KClO3 is 122.55 gm/mole)

= 0.30 moles

From the stoichiometry

10 moles of KClO3 is required to produce 3 moles of P4O10

when 0.30 moles of KClO3 is used x moles of P4O10 is formed

thus, 3 ÷ 10 = x ÷ 0.30

= 0.09 moles of KClO3 is produced

To know the mass of P4O10 apply the formula

mass = number of moles × atomic mass

= 0.09 × 283.886 ( atomic mass of P4O10 is 283.88 gram/mole)

= 25.55 grams of P4O10 could be produced.

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Rewritten by : Barada