1.) How many grams of aluminum sulfate, Al 2 (SO 4 ) 3, are needed to precipitat
ID: 1008529 • Letter: 1
Question
1.) How many grams of aluminum sulfate, Al2(SO4)3, are needed to precipitate 75.5 grams of solid calcium sulfate, CaSO4, in the following reaction. (Use Mole Ratio)
Al2(SO4)3(aq) + 3Ca(C2H3O2)2(aq) ---> 3CaSO4(s) + 2Al(C2H3O2)2(aq)
2.) How many grams of lead (II) sulfide, FeS, are produced when 150 mL of a 1.85 M ammonium sulfide, (NH4)2S, solution is mixed with 55 grams of iron (II) sulfate, FeSO4, solution in the following precipitation reaction. (Use Limiting Reagent)
(NH4)2S(aq) + FeSO4(aq) ----> FeS(s) + (NH4)2SO4(aq)
Explanation / Answer
Al2(SO4)3(aq) + 3Ca(C2H3O2)2(aq) ---> 3CaSO4(s) + 2Al(C2H3O2)2(aq)
From the balanced equation
3 moles of CaSo4 produced from 1 mole of Al2(SO4)3
3*136g of CaSo4 produced from 342g of Al2(SO4)3
75.5g of CaSO4 produced from = 342*75.5/3*136 = 63.3g of Al2(SO4)3 >>>>>> answer
2.
(NH4)2S(aq) + FeSO4(aq) ----> FeS(s) + (NH4)2SO4(aq)
no of moles of (NH4)2S = Molarity*volume in L
= 1.85*0.15 = 0.2775 moles
no of moles of FeSO4 = W/G.M.Wt = 55/152 = 0.36 moles
From balanced equation
1 moles of (NH4)2S react with1mole FeSO4
0.2775 moles of (NH4)2S react with 0.2775 moles of FeSO4
(NH4)2S is limiting reagent
1 moles of (NH4)2S to produced 1 mole of FeS
0.2775 moles of (NH4)2S to produced 0.2775 moles of FeS
mass of FeS = no of moles * gram molar mass
= 0.2775*88 = 24.42g of FeS >>>> answer
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