The following reaction can occur when a match is struck: KClO3(s) + 3 P4(s) ------- >...

The following reaction can occur when a match is struck:

KClO3(s) + 3 P4(s) ------- > 3 P4O10(s) + 10 KCl(s)

(a) How many grams of KCl(s) can be produced from 38.56 milligrams of P4 assuming that we have a sufficient amount of KClO3 present?

(b) In a separate experiment, how many molecules of P4O10(s) can be obtained from 0.00625 moles of KClO3? Assume we have a sufficient amount of P4.

Expert Solution

KClO3(s) + 3 P4(s) ------- > 3 P4O10(s) + 10 KCl(s)

3 moles of P4 react with KClO3 to form 10 moles of KCl

3*124g of P4 react with KClO3 to form 10*74.5g of KCl

38.56*10^-3 g of P4 react with KClO3 to form = 10*74*38.56*10^-3/3*124

= 28.5344/372 = 0.0767g of KCl

b. KClO3(s) + 3 P4(s) ------- > 3 P4O10(s) + 10 KCl(s)

1 mole of KClo3 react with P4 to gives 3 moles of P4O10

0.00625 moles of KClo3 react with P4 to gives = 3*0.00625/1 = 0.01875moles of P4O10

queen_honey_blossom answered 2 years ago

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