Question

6. 120.00 g of 0.00 g of Cadmium is allowed to react with 20.00 g of sulfur to produce cadmium (II) sulfide, How much of the product should you expect? (Use: Atomic mass of Cd 112.4 and S. 32 g. a) 2.7 g b) 27 g c) 40 g d) 20 g 7. If metal Zn is allowed to react with excess or enough sulfur and 30.7 grams of ZnS is obtained which represents 93.7% yield, then the theoretical yield should be a) 28.8 g b) 31.2 g c) 32. 89 d) 35.1 g e) 97.5 g 8. When 50.0 g of H2S reacts with 50.0 g of Oz, it will produce sulfur dioxide in the amount of (USE: 2 H,S+ 30, 250, +2H20 ) a) 100 g b) 67 g c) 2.0 moles d) 1.05 moles e) a and c f) band d 9. Define: Limiting reagent: % yield: 10. Which contains more number of atoms? a) 2.5 mole of CO b) 2.5 mole of CO2 c) 2.5 mole of H2O2 d) 2.5 moles of CCL

Answer 1

11214 112-4 112.4 100 Cd ts cds 112.4 gm of Cd reacts with 329 of Sulphug 20 gou 2 20 432 = 5.69 gm of sulphur but given 20 qm of sulphur limiting one = cdmium 112.4 go of ed~ Ke 1444 gm of cds 20gm of cd ? 20 X1444= molde od 227gm of eds | © en tres - Ins (5.37 65.379m of 20 - 30m0f2ms 65 Stross 91.37 32 Theoriticalyield = 93-7x30:7 = 28-7628.89 2H₂ S + 30₂ 2502 + 2H₂O 2 mol of Hy s perbal reacts with 3 moleseko 2x 349m of H2S reous with 3x329metom 50 gan " 2 50 m3X32 = 70 599m of o, requer 2x34 but given 50gr Limiting one is 02 3 mol of ₂ - 2502 3x32gmofoz 2x64 gm of soz 50gm 50 x 128 = 66.799 of sog. 96 267gm of eso, 9 kmiting: - The reasent which will be present in small quiantity decide the product formation 1. yield - It is the Ratio of Actual yield to Theoritical yield nucliplied by 100 10 2-6*6*023x16" x 2 = 15.075 x163.25*6-023x10x460-22x133 (b) 2-5%6:02 x 18 x3 = 45.1725x10 6.5 X6-023x10*3*5> 75 28x133 J 2.5 mol of cely