When solution containing 1.00 g of BaCl 2 and 1.00 gof Na 2 SO 4 were mixed, sol

Question

When solution containing 1.00 g of BaCl2 and 1.00 gof Na2SO4 were mixed, solid BaSO4formed. The BaSO4 precipitate was filtered out of thesolution, allowerd to dry, and then weighed. The mass ofBaSO4 that was isolated from solution was 1.40 g. Usethis information to answer the following question. Pay closeattention to the proper use and reporting of significant figuresand units. You must show all of your work (i.e., your calculations)for full credit. (a) Write a balanced chemical equation that shows the reactionthat occurs when an aqueous sodium chloride and solidBaSO4. In your equation, you should indicate thephysical states of each reactant and product. Note: Precipitationreactions are discussed in Section 4.2 of your textbook. (b) Calculate the moles of BaCl2 that are availablefor reaction. (c) Calculate the grams of BaSO4 that should formif all of the available BaCl2 reacts. (d) Calculate the moles of Na2SO4 thatare available for reaction. (e) Calculate the grams of BaSO4 that should formif all of the available Na2SO4 reacts. (f) Which reactant (BaCl2 orNa2SO4) is the limiting reagent in thisreaction? Why? (g) What is the theoretical yield of BaSO4? (SeeSection 3.10 of your textbook, pages 106-109, for informationregading the calculation of a percent yield.) (i) If this reaction went to completion, then how many gramsof the excess reagent should remain after the reaction iscomplete? When solution containing 1.00 g of BaCl2 and 1.00 gof Na2SO4 were mixed, solid BaSO4formed. The BaSO4 precipitate was filtered out of thesolution, allowerd to dry, and then weighed. The mass ofBaSO4 that was isolated from solution was 1.40 g. Usethis information to answer the following question. Pay closeattention to the proper use and reporting of significant figuresand units. You must show all of your work (i.e., your calculations)for full credit. (a) Write a balanced chemical equation that shows the reactionthat occurs when an aqueous sodium chloride and solidBaSO4. In your equation, you should indicate thephysical states of each reactant and product. Note: Precipitationreactions are discussed in Section 4.2 of your textbook. (b) Calculate the moles of BaCl2 that are availablefor reaction. (c) Calculate the grams of BaSO4 that should formif all of the available BaCl2 reacts. (d) Calculate the moles of Na2SO4 thatare available for reaction. (e) Calculate the grams of BaSO4 that should formif all of the available Na2SO4 reacts. (f) Which reactant (BaCl2 orNa2SO4) is the limiting reagent in thisreaction? Why? (g) What is the theoretical yield of BaSO4? (SeeSection 3.10 of your textbook, pages 106-109, for informationregading the calculation of a percent yield.) (i) If this reaction went to completion, then how many gramsof the excess reagent should remain after the reaction iscomplete?

Explanation / Answer

(a)     BaCl2 (aq)  + Na2SO4(aq) ------> BaSO4 (s) + 2 NaCl(aq) . (b)   Moles of BaCl2 = mass/ molar mass                                  =1.00 g/ 208.23 g/mol                                  =0.00480 moles . (c)   From the equation, we can see that onemole of barium chloride reacts to give 1 mole of bariumsulfate. So if all the barium chloride (0.00480 moles) reacts , weget 0.00480 moles of barium sulfate. . Mass of barium sulfate = moles * molar mass                                    =0.00480 moles * 233.39 g/mol                                    = 1.12g . (d)     Moles of Na2SO4 = mass/molar mass                                        =1.00 g / 142.04 g/mol                                        =0.00704 moles . (e)  From the equation, we can see that one moleof sodium sulfate reacts to give 1 mole of bariumsulfate. So if all the sodium sulfate (0.00704moles) reacts , we get 0.00704 moles of bariumsulfate. Mass of barium sulfate = moles * molar mass                                    =0.00704 moles * 233.39 g/mol                                    = 1.64g . (f)   From the masses of product obtained, weget lesser product when all the barium chloride reacts. Hencebarium chloride is the limiting reagent. . (g)   Theoretical yield is the mass of productthat is formed when all the limiting reagent reacts. When barium chloride(the limiting reagent) is completlyconsumed, we get 1.12 g of the product (as calculated in (c)). . (i)   When the reaction goes to completion, thelimiting reagent (barium chloride) is completely consumed. From theequation, we can see 1 mole of sodium sulfate (excess reagent)being consumed with 1 mole of barium chloride. . So we have 0.0048 moles of barium chloride being consumed and0.0048 moles of sodium sulfate are consumed along with it. Moles of sodium sulfate left over = 0.00704 - 0.0048                                                    =0.00224 moles . mass of sodium sulfate = moles * molar mass                                    =0.00224 moles * 142.04 g/mol                                   = 0.318 g So if all the sodium sulfate (0.00704moles) reacts , we get 0.00704 moles of bariumsulfate. Mass of barium sulfate = moles * molar mass                                    =0.00704 moles * 233.39 g/mol                                    = 1.64g . (f)   From the masses of product obtained, weget lesser product when all the barium chloride reacts. Hencebarium chloride is the limiting reagent. . (g)   Theoretical yield is the mass of productthat is formed when all the limiting reagent reacts. When barium chloride(the limiting reagent) is completlyconsumed, we get 1.12 g of the product (as calculated in (c)). . (i)   When the reaction goes to completion, thelimiting reagent (barium chloride) is completely consumed. From theequation, we can see 1 mole of sodium sulfate (excess reagent)being consumed with 1 mole of barium chloride. . So we have 0.0048 moles of barium chloride being consumed and0.0048 moles of sodium sulfate are consumed along with it. Moles of sodium sulfate left over = 0.00704 - 0.0048                                                    =0.00224 moles . mass of sodium sulfate = moles * molar mass                                    =0.00224 moles * 142.04 g/mol                                   = 0.318 g

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