Unit 6 - Stoichiometry & Chemical Formulas - Honors Addendum Flashcards
What is the Stoichiometric Rosetta Stone when answering Practice Problems?
1 mole = gfm = 6.022 x 10²³ Particles = 22.4L (g)
When given the number of moles, how do you find molecular weight, number of molecules/atoms and volume?
Gram Formula Mass - Given Number of Moles x Molar Mass
Number of Molecules- Given Num. of Moles x 6.022 x 10²³ Particles
Number of Atoms - Given Num. Moles x 6.022 x 10²³ x Num. Atoms of MOL
Volume - Given Num. of Moles x 22.4 L (g)
When given the number of particles, how do you find GFM, moles and volume?
Moles - GIven Num. of Particles/6.022 x 10²³
Gram Formula Mass - Moles x GFM of 1 Mole
Volume - Moles x 22.4 L
When given the molecular weight, how do you find number of particles, moles and volume?
Moles - Given GFM/GFM of one Mole
Number of Particles - Num. of Moles x 6.022 x 10²³ Particles
Volume - Num. of Moles x 22.4 L (g)
When given the volume, how do you find number of particles, moles and GFM?
Moles - Given Volume/22.4 L
Number of Particles - Num. of Moles x 6.022 x 10²³ Particles
Gram Formula Mass - Molar Mass x Num. of Moles
What is a trend seen all the time when solving for moles, number of particles, gram formula mass and volume?
- No matter what you are given at first,
- Number of moles must be found first
- To find all of the other values (GFM, Particles & Vol)
Definition of Limiting Reagent (2)
- Reactant that’s completely used up in a rxn
- Determines when the reaction stops
How do you calculate the Limiting Reagent? (7)
- Find GFM of the reactants (Don’t Include Coefficients)
- Given Grams/GFM = Moles
- Two Proportions
- Left Side: Coefficient Ratios of Reactants
- Right Side: X/Number of Moles Found
- Repeat for the Other One
- If one of the answers have more moles needed than the initial ones that you have then it is the Limiting Reagent
How do you calculate the amount of a product produced based on the Limiting Reagent? (3)
- Construct a Proportion between the Limiting Reagent & Product
- Left Side is the Mole - Mole Ratio from the Equation
- Right Side is X/Number of Moles of Limiting Reagent
How do you calculate the excess reagent left over based on the limiting reagent?
Initial Moles of Reactant - Moles Used = Excess Reagent
(To Calculate Moles Used, Make Sure the Ratio is 1:1)
Ex: Limiting Reagent is 2HCLO4 & Non is Ca (OH)2
2HCLO4 = 0.04 Moles
Ca (OH2) = 0.09 Moles
Ca (OH)2 (used up) = 0.04/2 (coefficient) -> 0.02 Moles
0.09 Moles - 0.02 Moles = 0.07 Moles
