Mole to Mole, G, L, & other Notes

Remember that you can get the following information from balanced equations: Reactants Products Moles MOLE RATIOS Molar/Formula Mass (using the periodic table)

New and Improved Mole Map A super bridge has been created that allows you to convert between different substances. Before you could convert between MASS, VOLUME, PARTICLES and MOLES…. Now you can convert from SUBSTANCE to SUBSTANCE using mole ratios.

Representative particles of G 1 mole G 6.02 x 1023 Representative particles of W 6.02 x 1023 1 mole W Mass of W mass W 1 mole W Mass of G 1 mole G mass G b mole W a mole G Mol G Mol W Volume of W (L @ stp) 22.4 L W 1 mole W Volume of G (L @ stp) 1 mole G 22.4 L G

2H20(l)  2H2(g) + O2(g) How many moles of O2 would you produce from 27.0 g H20? So..think we are starting with ____ of ____and we want to get to ____ of _____ Lets look at the chart! 27.0 g H2O O2 moles

1. How many moles of O2 would you produce from 27. 0 g H20 1. How many moles of O2 would you produce from 27.0 g H20 ? 2H20(l)  2H2(g) + O2(g) 27.0 g H20 x 1 mol H20 x 1 mol O2 = 0.75 mol O2 1 18.0 g H20 2 mol H20 Molar mass of H20 MOLE RATIO

Write what you start with Sn(s) + 2HF(g)  SnF2(s) + H2(g) How many liters of HF are needed to produce 9.40 L H2 at STP? Write what you start with Convert from L of H2 to moles H2 (1 mol = 22.4 L) Convert from mole H2 to moles HF (use mole ratio) Convert to liters of HF 9.40 L H2 x 1 mol H2 x 2 mol HF x 22.4 L HF = 18.8 L HF 1 22.4 L H2 1 mol H2 1 mol HF

Review Homework Questions? Now Practice! Do the Stoichiometry Worksheet B! You have 30 minutes then we will do Limiting Reagents

Limiting Reagents If you are working at a car factory and you have 20 frames and 40 tires, how many cars can you make? You have enough frames for 20 cars but only enough tires for 10 cars, so you can only make 10 cars. So the TIRES are the limiting factor..

Limiting Reagents Using the balanced equation you can determine the number of moles need to produce a specific amount of the product You can also determine which reactant you will “run out” of first – the limiting reagent

Example: 2H2(g) + O2 (g)  2H2O(l) Using the equation above, which reactant will be the limiting reactant if you have 4.0 g of H2 and 6.0 g of O2?

Covert to moles of H2 and O2 Compare moles using mole ratio 2H2(g) + O2 (g)  2H2O(l) Using the equation above, which reactant will be the limiting reactant if you have 4.0 g of H2 and 6.0 g of O2? Covert to moles of H2 and O2 Compare moles using mole ratio 4.0 g H2 x 1 mole H2 = 2.0 mol H2 2.0 g H2 6.0 g O2 x 1 mole O2 = 0.19 mol O2 32.0 g O2

Now you practice Problems on worksheet & then… Practice Problem on pg 313 Pg 321: #22

Percent Yield Percent yield = actual yield x 100 theoretical yield Actual yield is measured in the lab Theoretical yield is found using the balanced equation

Example: 2H2(g) + O2 (g)  2H2O(l) Using the equation above, if you started with1.0 L of H2 in the lab how many moles of water would you expect to make?

1.0 L H2 x 1 mol H2 x 2 mol H2O = 0.045 mol H2O 22.4 L H2 2 mol H2 2H2(g) + O2 (g)  2H2O(l) Using the equation above, if you started with 1.0 L of H2 in the lab how many moles of water would you expect to make? 1.0 L H2 x 1 mol H2 x 2 mol H2O = 0.045 mol H2O 22.4 L H2 2 mol H2 Lets go one step further…. Using the information calculated, what would be the percent yield if in the lab you measure 0.030 mol H2O?

Percent yield = actual yield x 100 theoretical yield Using the information calculated, what would be the percent yield if in the lab you measure 0.030 mol H2O? Percent yield = actual yield x 100 theoretical yield 0.030 mol H2O x 100 = 67% 0.045 mol H2O

What do I need to know? Mole ratio New and improved mole map (conversions) 1 mole = 22.4 L 1 mole = molar mass (g) 1 mole = 6.02x1023 particles Limiting reagents Percent yield UNITS!!! on everything