Question

Limiting Reactant and Theoretical yield Chapter 7 NAME Partners 4. LIOH + KCI → LiCl + KOH DATE ORROW LIA WORTlong guwollollo d Kai LCI + KOH bold stopalomarmor Mam H o w to log 0.81 marw action with 400 grams of lithium hydroxide. What is her theoretical yield of HOH a) Zuri began this reaction with 400 gram lithium chloride? b) Zuri actually produced 6 grams of lithium chloride. What is her percent yield o m bliy i9105 96 Yield - Actu Yleid - 100 Theoretical Yield 1) Balance the equation for the reaction of iron (III) phosphate with sodium sulfate to make iron (III) sulfate and sodium phosphate. F.PO4 + N. SO4 → Fo2(SO4)3 + Na PO W E a) If I perform this reaction with 25 grams of iron (III) phosphate and an excess of sodium sulfate, how many grams of iron (III) sulfate can I make? b) If 18.5 grams of iron (III) sulfate are actually made when I do this reaction, what is my percent yield?

Answer 1

To solve this problem, we must know what is theoretical Yield in this equation. The given reaction is Liot + ke lid + KOH i we see that the two reactants react in a lil mannen... So, the products are also formed in a 11 manner from the stoichiometry. hist. (a) zuni took 40.0 gm of Ligh- Molecular wt of LioH : 24.0 gm molts NO of moles of LOW : 40.0 moles: 50 There in a theoretical case žurios y M a noil Thus, in a theoretical case; } moles of Ligh yield exactly & moles of lied : Theoretical yield & Ex Moleculan wt of lid gm meticel yield3 1 14ms 2 x 42 gm du bolo i t i 1001 yilla (6) we see, zuni obtained 6 grams of lidt Thus ; 70 gm of lid corresponds to 1001 yield 6 gm of lid owl looxb. yieed 60g to yield 9.57. yield 2 = Hence; l. yield 8. 5.1 contrasten

only du balancing an eguangn, the number Yeau ant Bhul be lame on ball each the ma seaction arrow Fepog at Nag son real song + Naglia Na in present with a subranipt stoichiomely a on lihs and 12l on right hand side. Thus, the lemn of 2x3 is 2x3 = 6, we make wa 'b' in both sides. CASIC Step I : fe pou + 3 Nag sou fer (sou) + 2Nag Pou In the second and final step, if we just mun multiply repon by 2; the reaction is balanced Step - 26 epog + 3Nagson fer (sou) g + 2 Nag Pou . The correct balanced equation is ... 12 Fe pog+ 3 Naqson - fez (504) ₂ + 2 Nag Poal (a) from the balanced equation; we find repog + 3 Naqsoq h feq (504), & Naz Pog Thus ; . I mole of repog furnishes 12 moles of reason) kom gron (1) phosphate taken : 25 gms Molecular wt of gron (111) phosphate ~ 150 gm moll No of moles taken = 25 moles & 0 165 moles 6 0.165 moles of repog & hall furnish / x 0.165 moles of feq (sou) : 0.083 moles of Molecular weight of ferlson), fearsouza 400 gm molt Theoretical yield) : 0.083 x 400 = 133.11 gml (6) bercent yield = actual yield l using rogic of I previous calculation) . 118.5.x 100)), : 55.8%. 151 100%. oretical yield x 533 1 56 .