Question

Chemical Quantities and Aqueous Reactions 4. Determine the theoretical yield of H2S (in moles) if 32 mol Al2S3 and 32 mol H20 are reacted according to the following balanced reaction A1283(s) + 6 H2O() ? 2 Al(OH)3(s) + 3 H2S(g) A) 96 mol H2S B) 32 mol H2S C) 64 mol H2s D) 48 mol H2S E) 16 mol H2S

Answer 1

Calculating Numbeo Moles: . Mole:s is defined as the ya tid of the mass to the molar mass mass (m) Moles (n)? Molar mass (M) . Mass in Sram lor) Kilogvam Molar Hass in gram/mole or) kilosram/mo le 'Mass ỉs defined as the number of moles of the substance multiplied by the molar mass the substance Mass Cm) F moles (n) x Molar Mass (iM) Molar Mass is defined as the ratio of the the substante the mo les o mass (m) mo les (n) mass of the Subs tance to Molav Mass (M)

LIMITING REACTANT: The Limiting reactant i a chemical veact/oa is the Substance that totelily consumed when the chemical reaction is complete. The amount of Product formedis limi ted by this veagent, since the reaction cannot continue without Steps for solving limiting yeactant pyoblems: Step:1 »ees the Problem give you a mounts fov move thas one veactant? (at not. it's not a limitiag reactant Prob le m.) Step:2 Convert the amount of each veac tant to meles. step:31 рісколе of thereact ants. Use stoichiometry. to igure out how much of oae of the products you Could make f gous used this reactant 4P. tep : 4 :|Do the same for each of the othe> reactants tpThe reactant that can mane the least amount of Products gets used uP fivst. 5» it is the limiting veac tant . Step:6j LooK uf the number of moles ofPoduct that the limiting veactaat can make from steps 3 aad 4 step:7 rf the Problem asks howmuch of one oF the other veactants is left over, start with the moles o product that you can maxe o the limiting reactant (Step). Use Stolchiometry to figure out how much of this reac tant get used up. Substract this amount fron the number of moles you started with. Etep | 8:lıf the problem is asking for a ใแจก tity moles (such as gnms). convert the moles from other than Step 6 and Jor Step7 to the desred uit

Theoretical Yield: Theoretical Yieldis the uantity of product obtained from the complete convers/on of the limitng reacta nt in a reactant Chemical rea clion. t is the amount of Product resultinq from a perfect Chemical Yeactron and thus not the S ame as the you actually get &rom a reaction Theovetical Yreld S Commonly exp Presse d'o oF grams ormo les Actual Yield: Actual Yield is the amount of Product actually Produced by a react on. Actual vield is usually a smaller uantity because few ChemicaLveac tons pro ceed with 100% eFficiency, becau se of loss recovering the product, and because other reactionsmaųbe occur ring the Product. that redu ce sometimes actual yield is more than theoreticat yield, possiblu becausea second ary rea ction yield s product orbecause the recovered Contain s impurities

Percent Yield: .Percent ield is the percent vatio of actual vield to the thevretical ield .zt is calculated to be the experimental yield divided by the theoretical yield multiplied by 100%. the ·If the actual and theoretical yield are sam e , the Percent yield is100 %. Usually percent ield is lowe than 100y. because the actul vieid is often less then tie theoretical value. Reasos For this can include incomplete or compebn on competng. Yeactions and loss of sample during vecoverg. It's possible Fo percent yield to be over i007 w hich means more sample was recovered from a reaction than Predicted. This cahappen when othe v reactions were occurring tat also Formed the Product. can also be a Source of eror, if the excess is due to incomplete vemoval of wateγ or other impurities from the sanple. percent Yieldis always a Positive value Percent YieldActkal Yie ld X 100% Theoreti'cal Yield

Solut ion: Ca lcu late Theoretical Yield: 3 (s) C3) 2L) 3 2 moI. Moles 3 mol. HiS Mo les o 96 mo). H2S 3 mo I. H25 Moles oF HzS fom H,0 - 32mol.ho > 16 mol. H^ . Theovetical Yield o H S (in moles)