# Maximum amount of P4 [closed]

$\ce{2Ca3(PO4)2 +6SiO2 +10C->6CaSiO3 +P4 +10CO}$

What is the maximum amount of P4 that can be produced from 1.0 kg of phosphorite if the phosphorite sample is 75% $\ce{2Ca3(PO4)2}$ by mass

The wording is confusing. I do not know what to do. Would you start by finding the limiting reactant?

## closed as off-topic by MaxW, Jannis Andreska, Todd Minehardt, Jon Custer, Geoff HutchisonMar 20 '16 at 2:53

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First you find the number of moles of $\ce{Ca3(PO4)2}$ reacting. You have 75% of 1 kg, i.e. 750 grams, of calcium phosphate. The rest of reagents are assumed to be in excess as we have to find the maximum amount of $\ce{P4}$ obtained, hence reaction should not have other limiting reagents.