# Calcium Chloride and Oxygen

### A Guide to Solving Stoichiometry Problems

## Truc Tran

## What is stoichiometry?

Basically, stoichiometry allows us to convert and find the relationship between the reactant and its product in a chemical equation.

## Balanced Equation

Given the equation** Calcium chloride and oxygen.**

- Decide on the
**type of reaction**that is taking place. For calcium chloride and oxygen, the reaction would be synthesis. There is a rule that states*"Reaction of a metal chloride and oxygen produces a metal chlorate."***CaCl2 + O2 -> Ca(ClO3)2**or**Calcium chloride + Oxygen -> Calcium chlorate**should be your current equation. - Next step is to balance the equation. Since both sides already have the same amount of calcium and chlorine, all we need to do is balance the
**oxygen**. Simply add 3 in front of the oxygen on the**reactant**side to get six of it like in the product. The balanced equation should look like -**CaCl2 + 3O2 -> Ca(ClO3)2**

## Molar mass

**Mole to mole**or

**Mass to mass**conversions, it's handy to find the molar mass of the compounds ahead of time. You can get the mass by looking on the periodic table. Just add together the elements and multiply the elements that have a following subscript (Do not include coefficients). For example:

- Ca = 40.078 g
- Cl = 35.453 g
- CaCl2 = 40.078 + 35.453(2) = 110.984 g/mole

- O2 = 15.999(2) = 31.998 g/mole

- Ca(ClO3)2 = 40.078 + 2[35.453 + 15.999(3)] = 206.978 g/mole

## Mole to Mole Conversions

** Mole A (from given)| Mole B (Coefficient)**

** ----------------------------- | Mole A (Coefficient)**

Let's say that I got **3.18 moles of CaCl2**. This would be my given. So, let's go ahead and plug in the numbers to find **O2**:

__ 3.18 moles CaCl2|3 moles O2 __

-------------------------| 1 mole CaCl2

Multiply the top numbers and divide by the bottom to get the mole for O2. The answer will result in **9.54 moles of O2**.

## Mass to Mass Conversion

**mass.**There are more steps to this conversion, though it's pretty simple due to the fact that you only have to plug in the number and solve.

**Formula: **PT= Periodic table

__Mass A (given)|Mole A (always 1)|Mole B (coefficient)|Molar mass B(PT)__

---------------------|Molar mass A(PT) |Mole A (coefficient)|Mole B **(always 1)**

Let's say there's **12.1g of CaCl2**:

__12.1 g CaCl2| 1 mole CaCl2 | 1 Ca(ClO3)2 | 206.978g Ca(ClO3)2__ = **22.6 g of Ca(ClO3)2**

------------------|110.984gCaCl2|1 mole CaCl2|1 mole Ca(ClO3)2

## Limiting and Excess Reactant

__limiting reactant__

**limits**the amount of product that can be made.

__Excess reactant__ is the reactant that you have **extra** of.

To find the limiting and excess reactant, you have to do __mass to mass conversion.__

Problems that have **2 givens **are usually the ones asking for the limiting and excess reactants.

Let's say there are **12.3g of CaCl2 and 12.3g of O2** :

__12.3g CaCl2|1 mole CaCl2|1mole Ca(ClO3)2|206.978g Ca(ClO3)2__ =** 22.94 g Ca(ClO3)2**

----------------|110.984gCaCl2|1 mole CaCl2 | 1 mole Ca(ClO3)2

__12.3g O2| 1 mole O2| 1 Ca(ClO3)2| 206.978 g Ca(ClO3)2__ = **26.52g Ca(ClO3)2**

-------------|31.998g O2|3 mole O2 |1 mole Ca(ClO3)2

**CaCl2 is the limiting reactant and O2 is the excess reactant.**

## Theoretical yield and Percent yield

*t***The theoretical yield is the result of the limiting reactant when solved. CaCl2 was our limiting reactant, so**

*heoretical yield.***22.94 g**is the theoretical yield.

The amount actually produced is called the *actual yield*.

To get the *percent yield*, **divide the actual yield by theoretical yield then multiply by 100.**

Given the actual yield 25.01 g of Ca(ClO3)2 :

(25.01/22.94) x 100 = **109%**

__Reasons why the percent yield is not exactly 100%:__

- Contains impurities if its over 100%
- Less than 100% = Product is lost when transferring to a different container.
- Wrong measurements