Gas+Law+Teaching+Wiki+~+S.+Lindo

WELCOME TO MY GAS LAW TEACHING WIKI

My name is Sasha Lindo I'm a student in Dr. Reich's 422-004 Chemistry Class. In this wiki i will be providing you with some background information on four important chemists. I also will be teaching you about their gas law. Finding out the answer to some of the questions involved with gas laws equation you need to do stoichiometry. I will also be providing pictures and examples of the gas law and showing you how to solve a problem using a certain gas law. I will also have a page where you have do it yourself problems. There are two types of stoichometry: composition and reaction. Stoichiometry deals with the mass relationships of elements in compounds. Reaction stoichiometry involves the mass relationships between reactants and products in a chemical equation. Ex. If you had 824g of NH3 how many moles of NO would you produce in the equation NH3 + O2  à NO + H2O. 1st thing you would need to do is balance the equation. 2nd thing you would need to do is determine you’re given and you’re unknown. Then you would need to figure out how to get into NO from NH3. You start by getting out of grams of NH3 into moles of NO. You also need to find the molar mass of NO which is 17.04g. The balanced eqaution would look like: 4NH3 + 5O2 --> 4NO + 6H2O 824g NH3 X 1mol NH3/17.04g NH3 X 4mol NO/4mol NH3 = 48.4mol NO The g of NH3 cancel and then the mols of NH3 cancel leaveing you with the mols of NO and to go from mols of this same thing into the grams you just go backwards:

48.4mol NO X 4 Mol NH3 / 4mol NO X 17.04gNH3 / 1 Mol NH3 = 824g NH3 The mols of NO cancel out and then the mols of NH3 cancel leaveing you with grams of NH3. You could even take this question futher and turn it into millileter: 48.4mol NO X 4 Mol NH3 / 4mol NO X 17.04gNH3 / 1 Mol NH3 = 824g NH3 X 1ml X 1000ml =

Boyle's Law S.Lindo Charles Law S.Lindo Gay Lussac's Law S. Lindo Combined Gas Law S. Lindo Dalton's Law S. Lindo Ideal Gas Law S. Lindo Pratice Page S. Lindo Sites S.Lindo