# Solving Stoichiometry

### By: Alyson Nea

## Assigned Reaction

**Aluminum and Silver Sulfate.**

## 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**.

## Blanced Equation

**_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 i**s **6**, so we put **6 **next to **Ag** on the left side of the Equation.

## IUPAC Name

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**.

## Theoretical Yield

**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**.

## Precent Yield

**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**

## Proof Of Reaction

**sulfide and aluminum**takes place when the two are in contact while they are immersed in a

**baking soda solution**. The reaction is faster when the

**solution is warm**. The solution carries the sulfur from the silver to the aluminum.