Question

lab help please, ive answered a few questions need help on the rest

4. Consider the following problem and answer each question below to help you answer the overall questions posed here. This question is based on the reaction that you wrote in Question 3. A chemist allows 85.4 g of iron (II) chloride to react with 49.8g of hydrogen sulfide. How many grams of hydrochloric acid could be produced? How many moles of iron (II) chloride are present in the sample (10 points)? a. Mole of Fe Cl 0.527 mel giuot b. Based on your previous answers, how many moles of hydrochloric acid could be produced from 85.4 g of iron (I) chloride (10 points)? 2 mot of Fe Cl Produas 6 mol 0.529 mot af Fecle wdnces C. Based on your previous answers, how many grams of hydrochloric acid could be produced from 85.4 g of iron (Il) chloride (10 points)? mass of Hl = S8 max 86-5glmot d. How many grams of hydrochloric acid could be produced from 49.8 g of hydrogen sulfide (10 points)? mole of HoS44 g MOSS of H= 36 Tghudxa 929p 106.g 4647mot mote of Hcl-2x mote ef H2S 2X14647 2.929mot

Answer 1

solution Belance chemical Reaction:- 2Feclz + Hys —>2HC1+S +2 Fecha mass of iron Chloride (fccle) = 35.47 mass of Hydrogen sulfide = 49.89 (9) molar mass of iron Chosicke = 162.29 Imol so moles of Feclz in Sample= 85.49 - 162.2g/mol 0.52651 mol Anda b) according to balance Reaction 2 mole feclz Produce 2 mol HCI I mole fecla Produce 1 mol HCL 0152651 mol fedla Produce 0.52651 mol HCl So' moles of HCl = 0.52651 mole. Anne molar mass of HC1.= 36.46 g/mol moles of HCI = 0.5265) mole. (from part-6) So mass of HCl = 0.5265) PROX36467 -19.1969 mot mass of Hcl = 19:1969_Ans d) Mass of H2S = 49.89 moler mass of Hys= 34.1g/mol so moles of Hys = 49.87 146 mole. 341 g/mol . According to chemical Reaction I mol H₂5 Produce a mole holi 1.46 mol H₂S Produce 2x746 mol HU mole of HCl = 2.92 mol. so mass of HC1 = 2692 molx 36469 = 10646g_Ansa mol

(e) According to balance Reaction:- 2 mol fecla React with 1 mol HCl I mol fecha React with I mal. Hel 0.52451 mol fedla React with 4 xö•$2651 mol HCI -) 0.26325 mole Mole of Hcl in excess. So. Feclz is limiting Reactant. So Reaction Procedure depend on Ilimiting Reactanto Then maximum amount of MC1 = 19.1969 () Limiting Reactant is feclz Que 5 Balance chemical reactions. 2Mg (8) + 02 (9) 52mgo mass of magnesium = 2.200 (molar mass of mg = 24-30g/mol mass of oxygm gals. = 6:43g (molar mass of 0,= 160 90 /mol so' moles of magnesium= 2: 204 =) 0-09054 mole. 24.30 moles of magnesium oxide = 6-439 3) 0:2009 mole. Tog imor? According to balance Reaction 1 mole O React with 2 mole Mg 0.2009 mole Oz Reect with 240-2009 mole mg 0.4018 mole my. But Here 0.090 54 mole my is presente sui Jimiting React and is mg. Excess Reactent is Ozi According to Reaction ? mole my Produce 2 . more mgo 0.090 54 mole Produce 0.09054 mole ngo mote of mgo= 0.09054 mole, So Thoertical yield of Mgo = 40.30 gimol X0.09054 mo! =3•65g Anda According to Reaction; Reacted mole of O2 with 0.03051 Mg = 009054 mole XL -) 0.04527 mole. Reacted mass of 02 = 0.0452 mmole x32 H/mole >> 1459 on Reaction Compelete remain excess Reciclant Total mass of Oz - Realted mass (6.43-1.45); => 4:989 - Ans.