Question

13. b) Suppose only 4.02 grams of Fe(NO3), are recovered in the above reaction. What is the percent yield of Fe(NO), will form? 14. For the equation Pb(NO3)2 +K SO, à 2KNO, + PbSO, 22.9 grams of Pb(NO), is reached with an excess of K, SO, and 2.41 grams of KNO, are recovered. What is the percent yield? 15. Let's say the reaction H,SO, + 2 KOH à K.SO. + 2H,0, is known to proceed by a 20.85% yield via particular procedure. How many grams of KOH must be reacted with an excess of H,SO, in order to collect 67.5 grams of H.O? 16. For the equation 2C,H+90, à 6 CO, +6H20, 43.3 grams of C,H, is reacted with 52.0 grams 0. If the reaction proceeds via a 60.6% yield, how many grams of H.O will form?

Answer 1

actual yield x 100 7 Percent yield = - Theoretical yield Actual yield of KNO3 = 2.41g. Pb(NO3), + ga soy > I KNO3 + PbSoy Calculate theoretical yield of knių Pb(NO2 x Imole Pb(NO3) Imole kwaz 2x long = 13.98 g kno 1000 g 12.900 33112 g pool Imole Pb(NO3) Imole kno, 22 Percent yield = 2:41x100 - 17.24% 13.98 - H₂SO4 + 2 kon o soy + 240 Actual yield of the = 6705 g percent yield = actuel yield of the Theoretical yield 20185% = 67.5g 1 Theoretical yield Theoretical yield = 67. I x 10 = goram of tol must be reacted with an excess of thson. 323.709 10x hoc generen en het - 100719gog xon S 20.05 Imole kou

® 24467 9u2 - 6602 + 6140 - 43.39 of C346 ů reacted with 52 g of a Ist of all we find limiting reagent, one of the simplest way to identify a limiting reagent is to compare how much product each restant will produce. The reactant will produced less amount is called limiting reagent. using above equation. 43.39 9 4 6 x Imole (zhe x Gmole H₂O x 18 gho = 55.679 4o 42g 546 Imole ho imole yo 529 92 x Imole Iz x boole o logno = 19:59 40 - 32go quole q . Imole ho So, 19:59 of tho will be produced. (Theoretiel yield - Percent yield a . actual yield xko Theoretical yield 60.6 = actuel yield x 100 1905g actual yield = 60.681915 g = 11.8g. 11.8g of tho will form. ③ Reaction missing