Step 4: Stoichiometry of gases
Stoichiometry with gases everything stays the same but the equation you use to find the moles. It only takes basic knowledge of the gas laws to know why this is. Gas changes volume depending on temperature, and pressure. The relationship between volume and temperature is Charles's law which is at constant pressure a given mass of gas's volume increases as a factor of its temperature. Basically as it gets hotter its volume increases. The equation is V/T=k.
Charles's law must be done in Kelvin because it is an direct variation.
The other law is Boyle's law which shows the relationship between pressure and volume. The law states that at a constant temperature a give mass of gas's volume decreases as pressure goes up. The equation is PV=k
Gay-Lussac's law states the relationship between pressure and temperature. Again it must be done in Kelvin, since its a direct variation. In simple words it states as pressure increases so does temperature. the equation is P/T=k.
You can combine all three of those laws to get the combine gas law. Its equation is PV/T=k. 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 V/n=k
When all four of these laws are combined together they make the ideal gas law. The equation for that (this is the important part since we will be using it for stoichiometry) is PV=nRT.
R is the constant. The number changes depending on what units you measure it in. The basic outline of the units is PV/nT. I will be using the units kPa*L/n*K.
kPa- kilo Pascals
The n*K part will never change since the temperature must be measured in Kelvin and there are no other units for moles. The constant I am using is 8.314 L*kPa/n*K. To make it easier to solve stoichiometry problems arrange the equation
so it looks like this PV/8.314*K.
Now on to the actually stoichiometry. Nothing is really different just a different equation.
Here is an example using the reaction between oxygen and iron to produce ferric oxide.
You have 5L. of oxygen in a lab that is 300 Kelvin and 22kPa. How many grams of iron(III) oxide can you produce.
3 O2 + 4 Fe --> 2 Fe2O3
4.64g of ferric oxide can be produced
See exactly the same just using PV/RT.
Note that this will work for all gases, but if the gas is near is condensation point it will not work for it.