Question

The reaction between potassium chlorate and red phosphorus takes place when you strike a match on a matchbox. If you were to react 0.683 mol of potassium chlorate with excess red phosphorus, what mass, in grams, of tetraphosphorus decoxide could be produced?

Answer 1

To determine the mass of tetraphosphorus decoxide that can be produced when 0.683 mol of potassium chlorate reacts with excess red phosphorus, we need to use the stoichiometry of the balanced equation. The mass of tetraphosphorus decoxide that could be produced is 104.13 grams.

Step-by-step explanation:

The balanced equation for the reaction between potassium chlorate (KClO3) and red phosphorus (P4) is:

2 KClO3 → 2 KCl + 3 O2

To determine the mass of tetraphosphorus decoxide (P4O10) that can be produced when 0.683 mol of potassium chlorate reacts with excess red phosphorus, we need to use the stoichiometry of the balanced equation. From the equation, we can see that 2 moles of KClO3 react to produce 1 mole of P4O10. Therefore, to calculate the mass of P4O10, we need to convert the given moles of KClO3 to moles of P4O10 using the stoichiometric ratio.

Using the molar mass of KClO3 (122.55 g/mol) and the stoichiometric ratio from the balanced equation, we can calculate the mass of P4O10 as follows:

(0.683 mol KClO3) × (1 mol P4O10 / 2 mol KClO3) × (284.27 g/mol P4O10) = 104.13 g

Therefore, the mass of tetraphosphorus decoxide that could be produced is 104.13 grams.