Stoichiometry

Sodium Sulfate and Barium Hydroxide

Reaction

THE reaction is sodium sulfate and barium hydroxide. This specific reaction is a double replacement in which two elements which to create the product.

Na2(SO4) + Ba(OH)2~Na (OH) + Ba(SO4)

Balanced equation and IUPAC name

Balaced equation- 1Na2(SO4) + 1Ba(OH)2~2Na(OH) 1Ba(SO4)


IUPAC NAME- Sodium Sulfate + Barium Hydroxide~ Sodium Hydroxide + Barium Sulfate

Molar mass of each reactant and product

The molar mass is significant to finding the answer to many stoichiometry problems. This is found by multiplying the amount of elements by their mass (found on the periodic table of elements) and adding together the mass from the other elements In the reactants and products.


Ex: Sodium Sulfate/ Na2(SO4)

Na=22.990•2

S=32.066

O =15.999•4

(22.990•2)+32.066+(15.999•4)= 142.04

142.04= molar mass of sodium sulfate

Mole to mole conversion

Mole to mole conversions are the very simple. Set up with the given for the amount of moles in the first box. Then diagonally place the coefficient and product . Above that place the coefficient and product of what mole you want to switch to. Then multiply the top together and divide it by the bottom numbers.


Ex:1.26 moles of Na2(SO4)|1mole of Ba(OH)2= multiply these 2 numbers and divide by 1 mole of Na2(SO4)

Mass to mass conversion

Mass to mass conversions are a little bit longer but not very much more difficult. Start with your given in the first box,follow it diagonally with the molar mass of that particular product. Above that place the same product with just one mole. That's because under it is the molar mass of 1 single mole of the product. Next diagonally follow 1 mole with the coefficient of the same product. Above that put the coefficient and product of what kind of mass you're switching into. Next to that put the molar mass of 1 mole of that product and under that place 1 mole of that product since the molar mass is equal to 1 mole. Then multiply all the top numbers together and divide the top by each of the bottom numbers. Sounds confusing? Look at the picture/example,it will help as well as the website listed next to my real chemistry phone number.

Limiting and excess reactants + Theoretical yield

This is the same process as mass to mass conversions except there is 2 givens and each requires its own work. Both problems have to be worked out for the same reactant for a correct answer. Then the one with the smaller amount of the reactant is labeled as the Theoretical yield because, it's smaller and can't make anymore. The product that is used as a given to start for the theoretical is labeled as the limiting and the other product in the opposing problem is labeled as the excess, since it's the extra amount left over. Here is a website for another example- https://www.chem.tamu.edu/class/majors/tutorialnotefiles/limiting.htm

Percent yield

This is the easiest part of stoichiometry as long as you know the excess and limiting products. To find the percent yield all you have to do is divide the actual yield by the theoretical yield and multiply that by 100. This should range inverters 90-110%. Most often if it's off by a lot most likely something went wrong and wasn't accounted for in the expierement .


Ex: 8.03 divided by 9.04 • 100= 88.82%

Real world application

This reaction can be used to limit the amount of sulfur in the reactant sodium sulfate to create acid mine drainage (AMD) to creates helpful acid for mining any raw materials. This reaction typically has a yield of 85%.

If all else fails

If you've learned nothing from this here is a full 1 hour length guide to stoichiometry. Enjoy, and thanks for reading.


Parker McMillan

Mr.Miller

Aka dad

7th period preap chemistry