Question

1. The reaction of barium hydroxide with phosphoric acid produces an insoluble solid and water. When 10.00mL of 0.200M barium hydroxide solution is combined with 10.00mL of 3.00M phosphoric acid, 0.300g of the solid is produced.
2. What is the limiting reagent?
3. What is the theoretical yield of the solid?
4. What is the percent yield of the solid?
5. How much excess, in moles, is remaining?

Answer 1

3 Balott) + 2 HP 04 - Ba (Pula + 6H₂0 ) 192) laz) (s) Balott) molar mass= 171-34 glmol A Poy: Molar mass = 98 gimol B9₂ (Poul molar mass = 602 glmal Molarity of Balott - moles of Balot) x lodo volume (m) moles of Bacott), xlooo 0-2.2 - 10 moles of Ba(OH) = 0.002 Molarity of HPO4 - moles of H P 4 x 1000 volume (m) moles of HPO 4 xlooo 3 - 10 moles of H₂ P=0.03 From equation: Ralat react with 2 moles H₂PO4 0.002 moles salott), neact with a 9.902 * 2) moles H, por moles of HP required= 0.00133 moles But we have 0.03 moles H₂PO4 so ₂ Polis Excess reactant moles of Exceps readant remains = 0.03 0.00133 = 0.02867 moles

From equation: 2 moles HPO4 react with 3 moles Balott) 0.03 moles H Ro react with = 0.03 x3) moles Balott) moles of Balot required 0.045 moles But we have only 0.002 & moles of Balatt), so Balott, is zimiting reactant from equation: 3 moles. Balot) produce limole B9 (PO4)2 0.002 moles Ba(OH), produse = 0.60 2 x1) moles 89, (P2) Theoretical rield of zo.ooo 667 moles B9₂ (PO4), (inmoles) moles of mass of B9 (P) B9 (8) Molar mass of Bq (POL) mass of Bq (PO) 0.000667= 602 Mass of Bą (PO4) = 0.401534 gr (Theoretical wield) actual Weld xioo 0.3 x 100 y yield - - 74 - 71 . Theoretical 0.401534 yield