1.) How many grams of aluminum sulfate, Al 2 (SO 4 ) 3, are needed to precipitat

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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