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This question already has an answer here:

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

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


I am confused. How do I find a maximum amount of a reaction

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marked as duplicate by Jannis Andreska, Loong, Nicolau Saker Neto, Todd Minehardt, Jon Custer Mar 19 '16 at 23:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Phosphorite is a mineral containing calcium phosphate. This particular 1kg sample has 75% percent of it by weight i.e. 750 grams. All others reagents are assumed to be in excess as you have to find the maximum p4 obtained which would be at a point where all of our mineral is used.

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