By: Alyson Nea
Type of Reaction
Single Replacement. It's when a type of oxidation-reduction chemical react to an element or ion that moves out of one compound and into another.
A + BX -----> B + AX
Element A has replaced B (in the compound BX) to form a new compound AX and the free element B.
_Al + _Ag_(SO4)2 ------> _Ag + Al_(SO4)2
2Al + 3Ag2(SO4) ------> 6Ag + Al2(SO4)3
To Balance an Equation, both side must have the same Mass, But there are Coefficients that need to be moved. Since Al has switched places with Ag. The Coefficients between Al and (SO4)2 Must be changed. Al has 3 moles and it's mole has to switch with the Sulfate. Then, Al_(SO4)2 turns into Al2(SO4)3.
Now there are 2 Al's on the left side, and we have to balance the Al by multiplying 2 on the other side of the Equation. So, we place a 2 next to the Al. There's also 3 (SO4)2 on the left side of the Equation, so we have to put a 3 next to Ag to Balance out the Equation. Then there's 3Ag on the right, and since it's 3Ag2, it's multiplied by 2. 3 times 2 is 6, so we put 6 next to Ag on the left side of the Equation.
Aluminum + Silver Sulfate ------> Silver + Aluminum Sulfate
Because it's a Single Replacement, and Aluminum replaced Silver.
Molar Mass for Each Reactant and Product
Al + Ag(SO4)2 ----> Ag + Al(SO4)2
- Al = 26.982 g/Mole
- Ag = 107.87 g/Mole
- Ag(SO4)2 = 299.982 g/Mole
- Al(SO4)2 = 342.132 g/Mole
Mole to Mole Conversions
8.24| 1 mole Al(SO4)2 Multiplied together and then divided by |2 Mole Al to get 4.12 Mole Al(SO4)2 .
We add 1 Mole because it's the Coefficient from the Balanced Equation from Earlier. We Multiply it by 8.24 because It asks for my Birthday. Then, we Divided by 2Mole Al because it's also from the Coefficient from the Balanced Equation. After Multiplying and Dividing, We get 4.12 Mole of Al(S04)2.
Mass to Mass Conversions
12.1gAl X 1 Mole X 3 Mole Ag(SO4)2 X 299.992 = 10889.7096
10889.7096 Divided by 26.982 X 2 = 67.265g/Ag(So4)2
We started with 12.1 because it was the date. Then we Multiplied by 1 Mole because it's on top of a mass, and the mole is Always 1 if it's at the beginning. Then we Multiply by 3 Mole Ag(SO4)2 because it's the Coefficient from the Earlier Equation, then we multiply by 2999.992 because it's the Molar Mass of Ag(SO4)2. Then we get 10889.7069.
Afterwards we Divide by 26.982 because it's the Molar mass of Aluminum, then Multiply it by 2 because it's the Coefficient from the Earlier Equation.
Limiting And Excess Reactant
12.3g Al x 1 Mole Al X 6 Mole Ag X 107.868 g Ag X = 1174356.327
1174356.357 divided by 26.982 X 2 = 147.52g Ag
Limiting = Ag(SO4)2
Excess = Al
We started with 12.3 because it was the Date, then Multiply 1 mole because it's on top of a mass, and if a Mole Is on top of a mass, it's automatically 1. Then we multiply by 6 Mole Ag because It's from the Coefficient from the Earlier Equation. (6Ag + Al2(SO4)3). Then we multiply by 107.868g because it's the Molar Mass of Ag. Then we get 1174356.327, and we divide that with 26.982 because it's the Molar Mass of Aluminum. Then we Multiply by 2 because it's the Coefficient from the Equation. and we get 147.52g Ag. Our Limiting Element is Ag(SO4)2 and our Excess Element is Al.
12.3 Ag (So4)2 X 1 Mole x 6 Mole g Ag X 107.87 = 7960.806
7960.806 Divided by 299.82 g/Mole X 3mole x 1 Mole Ag =8.85g
We started with 12.3 because it was the date. We multiply it by 1 Mole because it's on top of a mass and it'd always start with 1. Then we Multiply with 6 mole Ag because it's a Coefficient. Then we Multiply by 107.87 because it's the Molar Mass of Ag. Then we get 7960.806. Then we divided by 299.82g Mole because it's the Molar Mass of Ag(SO4)2 and Multiply by 3 mole because it's a Coefficient and then we Multiply it by 1 mole Ag. And then our answer is 8.85g, which is Theoretical.
Actual Yield 6.99g Ag
So we put 8.85g to divide by 6.99 and then Multiply by 100 to get 126.60g/Mole