The reaction between potassium chlorate and red phosphorus takes place when you strike a match on a matchbox. if you were to react 65.1 g of potassium chlorate (kclo3) with excess red phosphorus, what mass of tetraphosphorus decoxide (p4o10) could be produced?

Since the problem states that excess amount of red phosphorus is supplied, therefore this only means that all of potassium chlorate are consumed to form the product tetraphosphorus decoxide.

The reaction is given as:

10KClO3(s) + 3P4(s) → 3P4O10(s) + 10KCl(s) 

First, we calculate for the amount of KClO3 in terms of moles. The molar mass of KClO3 is 122.55 g / mol.

moles KClO3 = 65.1 g / (122.55 g / mol)

moles KClO3 = 0.5312 mol

Using stoichiometric ratio of the balanced reaction, 10 moles of KClO3 produces 3 moles of P4O10, therefore:

moles P4O10 = 0.5312 mol KClO3 (3 moles of P4O10 / 10 moles of KClO3)

moles P4O10 = 0.16 mol

Converting this to mass by multiplying the molar mass of P4O10 = 283.886 g/mol

mass P4O10 = 0.16 mol * 283.886 g/mol

mass P4O10 = 45.24 g

ANSWER: 45.24 g P4O10