can someone help me with a tough chem problem?

Question

Ca3(PO4)2(s) + 5C(s) + 3SiO2(s) ---> 3CaSiO3(s) + 5CO(g) +2P(l)

An initial reaction mixture contains 1500kg calcium phosphate, 250kg carbon, and 1000kg SiO2.

After the reaction the slag was analized. It contained 3.8%C, 5.8%P, and 26.6%Ca by mass. What was the actual yield of phosphorus in Kg? what is the % yield?

(to save time: C is the limiting reactant and the theoretical yield of P is 257.8Kg, according to my calculations)

Answer 1

To save time, I'll go along with your limiting reagent, although the initial charge looks pretty close to equal. If the reaction was complete, there would be NO carbon in it.

Since the calcium is not going anywhere, you should be able to determine actual masses in the slag from the calcium [your initial mix has about 0.4 kg calcium/kg calcium phosphate or about 600 kg calcium ].
Then 600 kg is appx 1/4 of the slag mass or the slag mass is about 2300 kg. There is really two ways to go from here, and I'll work with the carbon balance (or inbalance ). Carbon in the slag is unreacted carbon . 3.8% corresponds to about 90 kg carbon. So the remaining 160 kg carbon must have reacted (which is now in CO). This is about 13 kg-moles, and from the reaction :
2/5 * 13 = 5.2 kg-moles of P formed.
From here, with the atomic weight of P (about 31 if I remember right), the rest is straightforward.