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Finding limiting reactant, theoretical yield, and actual yield

An electric furnace produces phosphorus by the following reaction:
Ca3(PO4)2(s) + C(s)+SiO2(s)= CaSiO3(s)+CO(g)+P(I)

An initial reaction mixture contains 1500g calcium phosphate, 250 g carbon and 1000g silicon dioxide.
a) What is the limiting reactant?
b) What is the theoretical yield of phosphorus?
c) After the reaction of the slag (solid residue) was analyzed. It contained 3.8 % Carbon, 5.8 % Phosprorus, and 26.6 % Calcium by mass. What was the actual yield of phosphorus? What was the percent yield?

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An electric furnace produces phosphorus by the following reaction:

Ca3(PO4)2( s) + C(s)+ SiO2 (s)= CaSiO3 (s)+ CO(g)+ P(I)

An initial reaction mixture contains 1500g calcium phosphate, 250 g carbon and 1000g silicon dioxide.

a) What is the limiting reactant?

This will be the reactant which is present in the smallest quantity, in terms of moles.

Formula Molar Mass Mass/g Moles (mass/molar mass)
Ca3(PO4)2 310.3 1500 1500/310.3 = 4.8
C 12 250 250/12 = ...

Solution Summary

This solution explains in 389 words how to calculate limiting reactant and yields using tables, calculations, and explanations.

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