In simple words it states as pressure increases so does temperature. P- pressure T- temperature k- constant You can combine all three of those laws to get the combine gas law. To use this with stoichiometry you need to combine it with one last law, which is Avogadro's law.
It states that at constant temperature and pressure equal volumes of gas contain the same number of moles.
The equation is PV=k P- pressure V- volume k- constant Gay-Lussac's law states the relationship between pressure and temperature.
Again it must be done in Kelvin, since its a direct variation.
I was looking around Instructables and saw many chemistry related Instructables, so I thought one on stoichiometry would help.
Basically stoichiometry (my definition) is the study of amounts in relation to a chemical reaction.This Instructable is good for people who are just learning stoichiometry and those who want an easier way. Now that you know the mole the second thing you need to do stoichiometry is balanced chemical equations. There are more complex equations that can't be balanced this way. Here we have the combustion reaction of sugar (which happens in your cells and powers your body).This is also great for all those people who like to use chemical reactions to blow stuff up. There are many different ways to find moles depending on what you are working with. C6H12O6 O2 -- H2O CO2 To balance this all you need to do is get the same number of atoms on both sides.Oh and one last thing once you have balanced the equation write it like this: 4 C3H7NO2 19 O2 -- 12 CO2 14 H2O 4 NO2 ^ ^ ^ ^ coefficient reacts produces chemical with symbol The basic outline: A, B, C, D- coefficients CHEM- chemical substance Sm- sample mass Mw- molecular weight n-moles A CHEM B CHEM -- C CHEM D CHEM Sm Sm l l Sm/Mw Mw*n l l n-----------------n/A=n/D----------------n Now to describe what this means. C6H12O6 6 O2 -- 6 H2O 6 CO2 26.3g 38.59g l l 26.3/180 44*.877 l l .146 moles-- .146/1 = .146*6 = .877 -- .877 moles 38.59g of CO2 will be produced This also works if you know the moles of one of the substances. Balance the equation and then find the moles and grams of oxygen produced if you have 3 moles of water.If you start with the mass of A CHEM then you divide the mass of that substance by its molecular weight(found using the periodic table) to find the moles. Some examples using the sugar combustion equation from the last step. H2O --- 2 H2 1 O2 x 48g l l x/18 32*1.5 l l 3 ----- (3/2)1 -----1.5 1.5moles 48g of oxygen produced Again remember to check your work.You can find the right amount to optimize the reaction and save reactants so you don't waste money. To help you understand how astronomically big this number is if I gave everyone on Earth (estimated 7 billion) million dollars a day; I could keep handing out money for 78564 years. For solids or liquids that aren't solutions its sample mass/molecular weight=moles. For example to find one mole of lets say carbon-14 the equation is x/14=1 x=14grams Another example: Find 5 moles of H2O. First off just pick a molecule(any of them work but the biggest is usually the best) and assign it a number (again any number works but to keep it easy use one).So far we have: C6H12O6 O2 -- H2O CO2 (1) So on the left side we have 6 carbon atoms 12 hydrogen atoms and an unknown amounts of oxygen atoms. Since we know the left sides number of carbon and hydrogen atoms we know the right sides number.6 CO2- Since we have 6 carbon atoms and each molecule has one atom in it we need 6 molecules of CO2 So now the equation looks like that and we are almost done: C6H12O6 O2 -- H2O CO2 (1) x (6) (6) Now that we know the number of molecules on the right we know the number of oxygen atoms in the equation.We have six H2O molecules each with one oxygen atom, so we have 6 oxygen atoms in those molecules.Now multiple the number of moles by the coefficient of substance you find the moles of and divide it by the coefficient of the substance you are translating it into. You can now find find the mass of the product produced using the Sm/Mw=n equation. You can use the law of conservation of mass to do this.Mass on both sides of the equation must be the same.