1. predict the precipitate that forms when aqueous solutions of silver nitrate and potassium chloride react to form products in a double-replacement reaction. include a discussion of how to write the complete chemical equation describing this reaction.

2. a 500-g sample of al2(so4)3 is reacted with 450 g of ca(oh)2. a total of 596 g of caso4 is produced. what is the limiting reagent in this reaction, and how many moles of excess reagent are unreacted?

al2(so4)3(aq)+3ca(oh)2(> 2al(oh)3(s)+3caso4(s)

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Its quite alot of work, but here it is
1. Silver Nitrate: AgNO3 (aq) 
Potassium Chloride: KCl 
To do a double replacement reaction, switch the two metals around, in this case Silver and Potassium so you are left with: 
Potassium Nitrate: KNO3 
Silver Chloride: AgCl 
The chemical equation should be something like this: 
AgNO3 (aq) + KCl => KNO3 + AgCl 

2The chemical reaction is as follows:Al2(SO4)3 + 3Ca(OH)2 → 3CaSO4 + 2Al(OH)3 Explanation:1mol of Al2(SO4)3 will react with 3mol Ca(OH)2 to produce 3mol CaSO4 and 2 mol Al(OH)3.First we have to find the number of moles of Al2(SO4)3 :Number of moles = Mass/ Molar massMass of Al2(SO4)3 = 500gMolar mass of Al2(SO4)3 = 342.15 g/molNumber of moles = 500/342.15Number of moles = 1.461 mol Al2(SO4)3Multiplying the coeffecient of Ca(OH)2 with 1.461:= 3*1.461 = 4.383 mol Ca(OH)2 Now we have to find the number of moles of Ca(OH)2:Mass of Ca(OH)2 = 450gMolar mass of Ca(OH)2 = 74.09 g/mol Number of moles = 450/74.09Number of moles =  6.074 mol Ca(OH)2We need 4.383mol to react completely with the Al2(SO4)3, so the Ca(OH)2 is in excess, and the Al2(SO4)3 is the limiting reactant. Excess unreacted: 6.074-4.383 = 1.69mol Ca(OH)2 unreacted 
 

Sorry i dont understand the question can you rephase it?