Limiting Reagent Calculations

Limiting reagent 3A+2B4C Limiting reagent calculations Balanced chemical equation is the “recipe” Amounts of reactants is how much can be made Limiting is which of A or B makes the least CPS: Limiting reagent CoCl2∙6H2O(aq) and Na3PO4(aq) precipitate Co3(PO4)2(s) Limiting reagent for mixtures, ACS page 103 (ACS 2.9) 3 Co2+(aq) + 2 PO43+(aq)  Co3(PO4)2(s) Net ionic equation: 3 Co2+(aq) + 2 PO43‐(aq)  Co3(PO4)2(s) Spectator ions: Na+(aq) and Cl‐(aq). colbalt + phosphate  precipitate Mixture 1: 0.15 M CoCl2∙6H2O + 0.15 M Na3PO4 Mixture 2: 0.15 M CoCl2∙6H2O + 0.075 M Na3PO4 Mixture 3: 0.075 M CoCl2∙6H2O + 0.15 M Na3PO4 Mixture 4: 0.075 M CoCl2∙6H2O + 0.075 M Na3PO4 Na+(aq) in 50. mL + 50. mL mixture 3, ACS page 103 (ACS 2.9) 3 Co2+(aq) + 2 PO43‐(aq)  Co3(PO4)2(s) Mixture 3: 0.075 M CoCl2∙6H2O + 0.15 M Na3PO4 Na+ comes from f Na3PO4(aq) ( ) and d so … 3 mol Na+ per mole of Na3PO4 Na+ is spectator and so … [Na+] = mol Na+/total volume What is the molarity of Na+(aq) after precipitation? Na+(aq) in 50. mL + 50. mL mixture 3, ACS page 103 (ACS 2.9) colbalt + more phosphate  same amount of precipitate  Co2+ limiting 3 Co2+(aq) + 2 PO43‐(aq)  Co3(PO4)2(s) Mixture 3: 0.075 M CoCl2∙6H2O + 0.15 M Na3PO4 moll Na N + = 0.15 mol Na3PO4/L x 0.050 L x 3 mol Na+/1 mol Na3PO4 = 0.0225 mol [Na+] = mol Na+/total volume = 0.0225 mol/(0.050 L + 0.050 L) = 0.225 M (sf?) Limiting reagent for mixtures 3 and 4, ACS page 103 (ACS 2.9) colbalt + phosphate  precipitate 3 Co2+(aq) + 2 PO43+(aq)  Co3(PO4)2(s) Mixture 3: 0.075 M colbalt + 0.15 M phosphate colbalt  0.075 M/3 = 0.025 M ppt phosphate  0.15 M/2 = 0.075 M ppt Mixture 4: 0.075 M colbalt + 0.075 M phosphate colbalt  0.075 M/3 = 0.025 M ppt phosphate  0.075 M/2 = 0.038 M ppt Why is mixture 2 pink but others are clear? Limiting reagent for mixtures 1 and 2, ACS page 103 (ACS 2.9) Limiting reagent for mixtures 1 and 2, ACS page 103 (ACS 2.9) 3 Co2+(aq) + 2 PO43+(aq)  Co3(PO4)2(s) 3 Co2+(aq) + 2 PO43+(aq)  Co3(PO4)2(s) Mixture 1: 0.15 M colbalt + 0.15 M phosphate colbalt  0.15 M/3 = 0.050 M ppt phosphate  0.15 M/2 = 0.075 M ppt Mixture 2: 0.15 M colbalt + 0.075 M phosphate colbalt  0.15 M/3 = 0.050 M ppt phosphate  0.075 M/2 = 0.038 M ppt Mixture 2: 0.15 M colbalt + 0.075 M phosphate colbalt  0.15 M/3 = 0.050 M ppt phosphate  0.075 M/2 = 0.038 M ppt What is [Co2+] after pptn? [Co2+] after pptn for mixture 2, ACS page 103 (ACS 2.9) 3 Co2+(aq) + 2 PO43+(aq)  Co3(PO4)2(s) start Co2+ = 0.15 M x 0.050 L = 0.0075 mol start PO43+ = 0.075 x 0.050 L = 0.00375 mol used Co2+ = start PO43+ x (3/2) = 0.005625 mol unused Co2+ = start Co2+  used Co2+ = 0.001875 mol [Co2+] = (0.001875 mol)/(.100 L) = 0.019 M Limiting reagent example, Dill/3e p. 12 7.39 kg of titanium dioxide reacts with excess carbon and chlorine according to the balanced reaction is TiO2 + 2 C + 2Cl2  TiCl4 + 2 CO. What is the minimum amount of carbon and chlorine that we need? The answer is 185 mol for each. Where does this number come from? g TiO2  mol TiO2  mol C, mol Cl2 Limiting reagent example, Dill/3e p. 12 Limiting reagent example, Dill/3e p. 15 7.39 kg of titanium dioxide reacts with excess carbon and chlorine according to the balanced reaction is TiO2 + 2 C + 2Cl2  TiCl4 + 2 CO. 3X g TiO2  mol TiO2: 7.39 x 103 g x (1 mol/79.88 g) = 92.5 mol TiO2 mol TiO2  mol C: 92.5 mol TiO2 x (2 mol C/1 mol TiO2) = 185 mol C mol TiO2  mol Cl2: 92.5 mol TiO2 x (2 mol Cl2/1 mol TiO2) = 185 mol Cl2 Limiting reagent example, Dill/3e p. 15 3X + 11.4 X  4Z  11.4 X x 4 Z/(3 X) = 15.2 Z 8.97 Y  8.97 Y x 4 Z/(2 Y) = 17.9 Y 2Y Y used = 11.4 X x 2 Y/(3 X) = 7.60 Y Y unused = Y initial – Y used = 8.97 mol – 7.60 mol = 1.37 mol + 2Y  4Z 11.4 mol X and 8.97 mol Y react. How much Z is formed? How much, if any, of X and Y remain?