Question

Q4 Solution Stoichiometry 10 Points A 25.0 ml of 0.527 M sodium sulfide(aq) and 55.0 ml of 0.243 Miron (III) nitrate are mixed together. What mass of solid iron (III) sulfide can be formed? What is the limiting reagent? Show your work for full credit. 3 Na2S (aq) + 2 Fe(NO3)3 (aq) ► Fe2S3 (s) + 6 NaNO3 (aq) The limiting reagent is Enter your answer here The theoretical yield of iron sulfide is Enter your answer here Please select file(s) Select file(s)

Answer 1

0.243 M = No. of moles of Fe(NO3)3 55/1000 L No. of moles of Fe (No,)3 = 0.243 x 55 1000 = 0.01336 moles calculation Limiting no: of moles of by This value will reagent To find the limiting reagent divide the compound, their respective stoichiometry. This be smaller limiting reagent. Let the Let this value be called x, and az for Ners and fe (Nos) respectively No. of moles of Nars 0.013175 feer x, = No. of = 0·00439 3 3 0.00668 X = No. of moler of Fe(NO3)3 20-01336 2 2 The value i the less far x, So Na, s in limiting reagent

of giver -0.91307 So all the calculatione will be done Nazs it with respect to less in amount 3 moles Na₂S I mole fez S3 I moles X0-013175 moles 207.9 has mass I mole molu 207.9 x 4.39x10 offerta off mole of Nas giver les of of fease 0.0 13175 moles of Nars Nags given giver 1/2 x = 4.39 x 103 moles offers FeS3 4:39 x103 mole of Fersz has mass = 207.9 x 4.39 x10 g ag of of g von (II) sulfide Therefore 0.9130 formed is in g of solid the given reaction with limiting reagent. a as Nos acting