# Reduction of chlorate with phosphorus

Given $$\ce{10KClO3 + 3P4 -> + 3P4O10 + 10KCl}$$

If 52.9 grams of $\ce{10KClO3}$ reacts with excess $\ce{P4}$, what mass of $\ce{3P4O10}$ could be produced? How many moles of $\ce{KCl}$ could be produced?

I am not sure if I have to calculate the excess mass of $\ce{P4}$ and then find the mass of $\ce{3P4O10}$ or if it is just telling me that $\ce{P4}$ is excess reactant. If it is calculating the excess mass, how would I do it if I do not know the original mass of $\ce{P4}$?

• When a question says Excess, it pretty much means "you don't need to worry about this chemical". The question is saying KClO3 is the Limiting reagent. – Jess L Mar 16 '16 at 19:10

So to solve this you determine the number of moles of $\ce{KClO3}$ and use that as the base for determining the amount of $\ce{P4O10}$ and the moles of $\ce{KCl}$ produced. Then multiple by the molecular weight per usual to get your answer.