Chemistry 11 - Limiting Reagent question

2 Ca3(PO4)2(s) + 6 SiO2(g) + 10 C(s) --> P4(s) + 6 CaSiO3(s) + 10 CO(g)
If 39.3 g of Ca3(PO4)2, 24.4 g of SiO2 and 8.00g of C are available, find the limiting reagent.

My final answer is Ca3(PO4)2 is the limiting reagent. Is this correct? Thanks.

Yes your answer is correct knowing that the number of moles of Ca3(PO4)2 , SiO2 , and C are approximately equal.
n Ca3(PO4)2= 0.127 moles
n SiO2= 0.135 moles
n C= 0.133 moles

find moles of each of them individually then divide by their coefficent. the answer which is the lowest is your limiting reagent.

moles of Ca3(PO4)2 divide by 2 =

moles of Sio2 divide by 6 =

moles of c =10 =

lowest answer is LR