Reduction of chlorate with phosphorus Given 
$$\ce{10KClO3 + 3P4 -&gt; + 3P4O10 + 10KCl}$$

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  $\ce{P4}$?
The number of moles produced for each product will be the number of moles times of the limiting reagent times the reaction stoichiometry (the coefficients).

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.