# Guide to Stoich

### KCl+O

## Intro to Chemistry

## Balancing the equation

**2KCl+3O2->2k(ClO3)**

We want the same as everything all around so we add numbers before elements, subscripts, to multiply them with the number under the element to balance it out. In this case the product becomes i metal chlorate.

I know it's a lot to take in, so keep up with me, it'll make sense in the end.

## Name for each reactant and product

__Potassium Chloride Plus Oxygen Produces Potassium Chlorate__

You write down what you see but in words

## Here's a cool little video so you get more interested in this stuff

## Molar Mass for each reactant and product

**2KCl+3O2->2K(ClO3)**

Let's start by splitting it into smaller sections, we get molar mass from the periodic table, so with each element put the molar mass from the periodic table, as so.

**K-39.098(2)+ Cl-35.453(2):** this is one molar mass, then you move to the next one.*(149.102g)*

**O-15.9994(6):** This is the second part, now we move to the product.*(95.9964 g)*

**K-39.098(2)+Cl-35.453+O-15.9994(3):** this is the product*(161.6472 g)*

Now we have the molar mass for everything and you can answer whatever the question may be, they may ask for product or reactant molar mass, you're prepared to answer.

## Mol to Mol conversions

When doing stuff in chemistry you want to have both your elements common, in this case it's grams.

You get __Mol A __and Multiply it by **Mol B**

Once you get that you divide it all by __Mol A __as so. (Note) __Mol A __is always given

11.30 mol KCl* 2K(ClO3)

------------------------------------

2KCl

That gives you 11.3 g K(ClO3)

Use the Coefficients infront of* *__Mol A__ and **Mol B** that are given in the equation. In this case we're given 2 and 2 on both __Mol A__ and **Mol B**, those can cancel out if you wish to do so.

## We still have much more to do, but we're almost done and you're doing great

## Mass to Mass Coversions

__Mol A __(Given)* __Mol A __(Always 1)***Mol B **(remember the coefficients from the original equation)* Molar mass of **Mol B **(from periodic table)

this all divides by these next steps

Molar mass of __Mol A__ and __Mol A__(don't forget the coefficient) and **Mol B** (always 1)

Seeing it will help you a lot so lets do that.

8.19K(ClO3)*1 mol K(ClO3)*3O2*31.9988O2

---------------------------------------------------------------

122.5442K(ClO3)*2K(ClO3)*1 mol O2

Note: I got 8.19 given to me.

After all the math we get: **3.21 g K(ClO3)**

## Limiting and Excess Reactant

5.25K(ClO3)*1 mol K(ClO3)*3O2*31.9988O2

---------------------------------------------------------------= **2.06 g K(ClO3)**

122.5492gK(ClO3)*2K(ClO3)*1 mol O2

Now we solve for oxygen so we can figure out which one is excess and which one is limiting

12.18O2*1 mol O2*2K(ClO3)*1.22.5492gK(ClO3)

----------------------------------------------------------------------= **31.11 g O2**

31.9988gO2*3O2*1 mol K(ClO3)

Now that we've calculated both of them we know that the smaller number is the Limitng reactant

## Theorectical Yield

2.06

---- *100

2.06