Question

Question 3 In a laboratory experiment, 23.4 g of Cu were reacted with 141.9 g of HNO, 3Cu + 8HNO, — 3Cu(NO3)2 + 4H20 + 2NO (a) How many grams of Cu(NO3)2 can be theoretically produced? g Cu(NO3)2 Use correct number of significant digits, the tolerance is +/-2% (b) How many grams of the excess reactant remain? Excess reactant: Edit Grams remaining Use correct number ondant as the verge 5 8

Answer 1

Answer 3 Cy (NO3)2 + 4420 + 2NO muss molar mas 3c4 + 23.49 63.546 glome 8 Hnog — 141-99 63.01g|mo No. of moles of C4 = 23.44. 63.546gimol 2 0.3682 mol No.of moles of Hwag = 141.98 63.018/mo, = 2:2520 moi 03.018/mo. - 3 mol cu react with 8 mol HNO3 so 0.3682 mol cu will react with fx Q. 3682 mol = 0.9819 mol Hnog means HNO3 is excers limiting reactant. reactant and ce is from so Amount limiting reactant of Cu(NO3)2 produced will depend on moles of an (nogle produced - holes of cu reacted noles of Cu (Na la produced = 0.3682 moj Amount of Cu Nogla produced = = No. of moles & Molur mas 0.3682 101 X 187.56 g/mol 169.06 g

D moles of excess HNO3 = (2.2520_ 0.9819) = 1.2701 mol Amount of excess HNO3 = molex molar mass = 1.2701 mol x 63.0181mol = 180.03 g % yield = astuced yield Xloo theoreticalyield 71.9 = acuધુ થવા - X 100 69.06g | Actual yield = 49.65 g