Question

The reaction between potassium chlorate and red phosphorus takes place when you strike a match on a matchbox. This happens in the reaction

KClO3 + P4 = P4O10 + KCl

If there are 0.895 g of KClO3 and 0.995 of P4
a. Which is the limiting reactant
b. How much KCl is produced in the reaction?

Answer 1

3 P₄ + 10 KClO₃ → 3 P₄O₁₀ + 10 KCl

P₄ is a reducing agent, KClO₃ is an oxidizing agent.

phosphorus is the limiting reactant

Moles of $KClO_3=\frac{0.895}{122.55}=0.0073moles$

Moles of $P_4=\frac{0.995}{124}=0.0080moles$

Moles of $KCl= \frac{10}{3}×0.0080=0.027moles$