Question

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(aq)--->2Al(OH)3(s)+3CaSO4(s)

Show all work please.

Answer 1

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

Answer 2

Answer 1:

Step-by-step explanation:

Double displacement reaction: It is a reaction in which reactants exchange their ions to form products in chemical reaction.

`Ab+Cd\rightarrow Ad+Cb`

When aqueous solutions of silver nitrate and potassium chloride reacts together to form white precipitate of silver chloride with potassium nitrate in aqueous solution. While writing chemical reaction, first write the molecular formula of silver nitrate and potassium chloride with 'addition' sign in between(+) on the left hand side. Followed by right arrow and then the molecular formula of products formed on the right hand side

`AgNO_3+KCl\rightarrow AgCl+KNO_3`

Answer 2: 1.69 moles of
`Ca(OH)_2` remained unreacted.

Step-by-step explanation:

`Al_2(SO4)_3(aq)+3Ca(OH)_2(aq)\rightarrow 2Al(OH)_3(s)+3CaSO_4(s)`

Number of moles of
`Al_2(SO4)_3`=

`(500 g)/(342.15 g/mol)=1.46 moles`

Number of moles of
`Ca(OH)_2`=

`(450 g)/(74.09 g/mol)=6.07 moles`

According to reaction , 1 mole of
`Al_2(SO4)_3` react with 3 moles of
`Ca(OH)_2` , then 1.46 moles of
`Al_2(SO4)_3` will react with :
`1.46* 3` moles of
`Ca(OH)_2` that is 4.38 moles.

1.46 moles
`Al_2(SO4)_3` react with 4.38 moles of
`Ca(OH)_2`. So, this means that
`Al_2(SO4)_3` is present as limiting reagent. And
`Ca(OH)_2` is an excessive reagent.

Moles of
`Ca(OH)_2` left unreacted :6.07 moles - 4.38 moles =1.69 moles

Hence,1.69 moles of
`Ca(OH)_2` remained unreacted.