What is limiting reactant in chemistry?

Help me with this Chemistry problem? (% yield, limiting reactant)?

• 4.477x10^2 molecules of Ca3(PO4)2 are mixed with 0.1557g of SiO2 and 0.02783g of carbon solid. What is the percent yield if 0.05942g of CO are actually obtained in the lab? 2 Ca3(PO4)2(s) + 6 SiO2(s) + 10 C(s) → 6 CaSiO3(s) + P4(s) + 10 CO(g)? First off, there are THREE reactants. How the HECK do I find the limiting reagent from THREE reactants?! I don't know. Do I NEED the limiting reagent (or limiting reactant)? I know I need to find the theoretical mass of CO, but HOW do I do it?! PLEASE HELP ME! I'm SO confused, and I have a test on this TOMORROW! I BEG YOU!

• Answer:

  Hi Sidney, 2 Ca3(PO4)2 + 6 SiO2 + 10 C → 6 CaSiO3 + P4 + 10 CO I think there is an error . . . ! I think that " 4,477•10^2 molecules " is in reality 4,477•10^-2 mol of Ca3(PO4)2 . . . so : LIMITING REACTANT . . . ! ¯¯¯¯¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯ You calculate the number of mol of reagents . . . n(Ca3(PO4)2) = 4,477x10^-2 mol n(SiO2) = m(SiO2) / M(SiO2) n(SiO2) = 0,1557 / 60 n(SiO2) = 2,595•10^-3 mol of SiO2 n(C) = m(C) / M(C) n(C) = 0,02783 / 12 n(C) = 2,32•10^-3 mol of C . . . then : You calculate the constants associated with the reagents by dividing the number of moles of reactants by the stoichiometric coefficients . . . Ca3(PO4)2 : k1 = n(Ca3(PO4)2) / 2 . . . . . . . . . . . . k1 = 4,477x10^-2 / 2 = 2,234 x10^-2 SiO2 : k2 = n(SiO2) / 6 . . . . . . k2 = 2,595•10^-3 / 6 = 4,325•10^-4 C : k3 = n(C) / 10 . . . k3 = 2,32•10^-3 / 10 = 2,32•10^-4 . . . then, you compared the constants . . . ! The smallest constants indicates... limiting reagent ! k3 < k2 < k1 . . . then C is the limiting reagent . . . ! % YIELD . . . ! ¯¯¯¯¯¯¯¯¯¯ . . . according to the equation : 10 mol of C gives 10 mol of CO . . . therefore : 2,32•10^-3 mol of C gives 2,32•10^-3 mol of CO . . . ! Theoric mass : m(CO) = M(CO) x n(CO) m(CO) = 28 x 2,32•10^-3 m(CO) = 0,06494 g of C % yield = 100 x m(real) / m(theoric) % yield = 100 x 0,05942 / 0,06494 % yield = 91,5 %