Question

) sulfide reacts with nitric acid according to the reaction below: 5. Antimony(III) sulfide reacts w Sb2S3(s) + 10 HNO3(aq) → Sb2Os(5) a. Calculate the theoretical yield in sulfide reacts with 0.579 mol nitric acid. O HNO3(aq) Sb Os(s) + 10 NO(g) + 3 S(s) + 5 H200 clical yield (in grams) of each product when 32.1 g of antimony(1) 321 gs bz63 x 1 mol sbg 82 x mol sbag Molsb2S₃ 0.519 HNO3 x 01 Slezs - Imolsberg Omo IVO, inolo629 = b. How much of the excess reactant (in grams) is consumed in the reaction? 6. a. If 0.405 mol butane (C4H10) and 53.9 g oxygen react, how many moles of water will be formed? 2 CHg) + 13 02(g) → 8 CO2(g) + 10 H2O(g) vield. What was the actual yield of carbon b. The reaction above was found to proceed in 89.7% yield. What was the dioxide isolated?

Answer 1

Date Page Sb₂S3 + 10 HNO Sbzos & IONO, + 35 +5H₂O 1 339.7 gimol mol of Sbez sreached = 32.1 g - ologun male of No = 0.579 mo. male She Sz 10 mol HNO, nmol 1 - 0.579 mole Hinog 2 = 0.0579 moh (But s baz is 0.0944) Sb₂Sa in exces HNO3 is limiting reagent 0.0579 mol l Sbos ywered = strane mals " " = 0.0579 moly 323. 57 simot " " = 18.73 am I No. 600 o.579 mole mas of No₂ = 0.575 modex 465/m = 26.639 S = 0.173 mol turno mars og s= 0.173 moche 329 / mot = 5.56 g make of to formed a sx o. 579 = 0.289 mar os te farmed -0.289 molx is si mot = 5.21 Gram

/ Date Page RANKA Exess Shas left = 0.0944 -0.05791 = 0.0365 moh. mals of Sb, s, (toht) = 0.0365 mol x 339. 71/mt = 12.39 gram