Question

check answer please

10. 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.

n Ca3(PO4)2= 0.127 moles
n SiO2= 0.135 moles
n C= 0.133 moles

My final answer is Ca3(PO4)2 is the limiting reagent.



If the conversion of Iron ore to iron is represented by Fe2O3(s) + 3CO(g)  2Fe(s) + 3CO2(g) produces 147 g of iron from 500 g of iron ore, what is the percent yield?

Fe2O3(s) + 3CO(g) --> 2Fe(s) + 3CO2(g)
that tells you 1 mole Fe2O3 = 160g gives 2 moles Fe =112g

Use simple proportion
500g Fe2O3 gives 112/160 x 500= 350g

% yield = actual amount / calculated amount x 100 = 147/350 x 100 = 42%

Answer 1

I don't agree with the limiting reagent I get different values for mols SiO2 and mols C. I agree with mols Ca3(PO4)2.

I agree with the 42$ percent yield.

Answer 2

Convert your given mass values to moles then divide by the respective coefficients of the balanced equation. The smallest value is the limiting reactant.