Question

5. A student performs the following reaction using the procedure outlined in Part A of this experiment. Ba(NO3)2 (aq) + K2SO4 (aq) - BaSO4(s) + 2 KNO3(aq) The student adds 55.0 mL of 0.050 M K SO. solution to 50.0 ml of 0.050 M Ba(NO3)2 solution. The precipitate of Baso. is collected by gravity filtration and dried to a constant mass. The BaSO.is determined to have a mass of 0.49 g. (See section 3.9 in the textbook for a review of stoichiometry). a. Determine the limiting reactant for the reaction performed by this student. (2 pts) b. Determine the theoretical yield of the BaSO4 precipitate. (2 pts) c. Determine the %yield of BaSO. (1 pt)

Answer 1

5. Given reaction, Now Ba(NO3)2 (aq) + 6,504 (aq) + Basoy (S) + 2 KNO3(aq) 55.0 ml of 0.050 kson = 0.050 × 55.0 mol af ksoy = (2.75 x 10-3 mol of K,son 100 50.0 ml of 0.050 m Ba(NO3), 0.050 X 50,0 1000 mal af Balmo a for limiting =(2.5 x 10-3 mol of Ba(NO3). reagent, (n.) value is low.

here, n= n af moles sie - stoitiometric ce-efficient 2.5810-3 (sie) BalNo = 23x10 = 2,5x10-3 d) xson - 2:35x10-22-75*10*? So, here Ba(NO3)₂ be the limiting reagent. the balanced reaction, from I mole of Ba(NO3)₂ we get 1 mole of Basou from (2.5x103) mal of Ba(NO3), we get (2.5 * 18-3) mel of Bason Now, molecular weight of Basoy is 233.38 g/mol : (2.5x10-3 mol of Basoy = (2.5x233.38×10 3 g of Basoy - 0.588 & Bassy b. from Hence, theoritical yield of the Boso y precipitate will be 0.58 g. e. Given experimental yield = 0.49 g. we get theoritical yield = 0.589 experimental yield : Percent of yield = theoritical yield X160 y. = 84.484 Hence, the percent of yield will be 84.48%.