Question

Aluminum hydroxide reacts with sulfuric acid as follows: 2Al(OH)3(s)+3H2SO4(aq)→Al2(SO4)3(aq)+6H2O(l)

a. Which reagent is the limiting reactant when 0.450 mol Al(OH)3 and 0.450 mol H2SO4 are allowed to react?

b. How many moles of Al2(SO4)3 can form under these conditions?

c. How many moles of the excess reactant remain after the completion of the reaction?

Answer 1

du balance Ran adl (on)(s) 3,Se, (ae)-dl (sa) (a)+ 6M,0() Stehford Hiiting ege a) mele fue loff of MllOn mals Calfidenot b male hu lotf of , soy=0.45 O.45-0-225 2 O.15 mals hue loff f somale h.C of Allon LinitHoy Pengut steh applieol POAC Rule for babnted Ra males of M, Soy Caffident moles of ds (sou) meles of dls Csan) Cofficient 0.45 3 0.15 mol males

tehird exes males of dl (o ogain applied Po AC ule males of Alons Cofficent males of Ha.50y ofaient moles of Allon)s O.45 2 O.45x2= 0.3 mot moles of dlon)s So mols of dllOh), Corallumed in oxh= 0.3mol. Remaining males of oh initial malslondumed meles = 0.45-0.3 =0.15mal meles of dllon), Remains = 0.15 mal 3 Thank Jau, May it helf yau