Guide to Stoich

KCl+O

Assigned Reaction

Potassium Chloride+ Oxygen

Type of Reaction: Synthesis

Intro to Chemistry

Welcome to chemistry, where you think you know something but your dreams get crushed. I'll be your guide, Abdullah, I'll make it as simple as i can, come along this ride with me.

Balancing the equation

Balancing sounds complicated but it really isnt. You want the same number of elements on the reactants side as the products side. As our example states.

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

This is when we basically just name it out as so.

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

Potassium Chlorate Vs 5lb Gummy Bear

Molar Mass for each reactant and product

Sounds complicated right? Trust me, it's not. I thought i'd never get this stuff, now i do it with ease. Lets start by writing down our equation

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

Mol to Mol is confusing but we can get it.

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

Much like the prior step, we have to get the mass of elements to solve, so thats what we're doing. It goes like this:

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

When calculating what we can make we're going to have more of a certain compound, thats called the Ecess, the other is the limiting, the lower number, the one that limits how much product you can make. To figure this out we have to figure out how much we have for both elements by doing exactly what we did in the step before, but this time, the number thats given to us is changed to 5.25, which will alter our calculations, we're also figuring out the mass of oxygen, let's get started.

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

You divide what you got from your expierement and your limiting reactant then multiply by 100 as so.

2.06

---- *100

2.06

Percent Yield

This would equal a 100 percent which would mean we did our calculations perfect.

Conclusion

Thank you for taking this trip with me. I hope you learned a lot from this and i hope you understand chemistry well, now before you leave go watch that video again because it was cool. Have a great year!