Stoichiometric Calculations Foldable

Foldable Obtain 3 sheets of paper. Stagger the pages about ¾ of an inch. Fold over so you have multiple flaps.

Titles of Pages Stoichiometric Calculations Mole to Mole Mole to Mass / Mass to Mole Mass to Mass Limiting Reactant Percent Yield

Page 1 - Top Open your book, on the top flap, write the following: Steps in all stoichiometric calculations: Write a balanced equation Identify the known and unknown, what do you have and what are you trying to find? Set up the problem with the given information Identify the mole ratio Check your answer, does it make sense?

Page 1 – Bottom – Mole to Mole Mole of A to Mole of B Ex. How many moles of H2O are produced from 5.0 moles of H2? Balanced Eq: 2H2 + O2  2H2O Known: 5.0 mol H2, Unknown: mols of H2O = 5.00 mol H2O 5.0 mol H2 2 mol H2O 2 mol H2 Mole Ratio

Page 2- Bottom – Mass to Mole Mole of A to Mass of B Mass of A to Mole of B How many grams of NaCl are produced from 5 moles of Na? Balanced Eq: 2 Na + Cl2  2 NaCl Known: 5 moles of Na Unknown: grams of NaCl How many moles of NaCl are produced from 35 g of Cl2? Balanced Eq: 2 Na + Cl2  2 NaCl Known: 35 g Cl2 Unknown: moles NaCl 5 mol Na 2 mol NaCl 58.44 g NaCl 35 g Cl2 1 mole Cl2 2 mol NaCl 2 mol Na 1 mol NaCl 70.90 g Cl2 1 mole Cl2 Mole Ratio = 0.99 mol NaCl = 292.20 g NaCl

Page 3 – Bottom- Mass to Mass Mass of A to Mass of B Ex. How many grams of NO2 are produced from 250 g of N2 gas? Balanced Eq: N2 + 2O2  2NO2 Known: 250 g of N2 Unknown: grams of NO2 = 821.02 g NO2 46.01 g NO2 250 g N2 1 mol N2 2 mol NO2 28.02 g N2 1 mol N2 1 mol NO2 Molar Mass Mole Ratio

Page 4 – Bottom – Limiting Reactants Same as all the other problems, you just have to do two of them. If you have 25 g of N2 and 35 g of O2, how many grams of NO2 is produced, which one is the limiting reactant? The excess reactant? Do the problem with N2 first. Balanced Eq: N2 + 2O2  2NO2 Known: 25 g N2, Unknown: g NO2 = 82.10 g NO2 Now do the same problem with O2 Same balanced equation Known: 35 g O2 ,Unknown: g of NO2 = 50.32 g NO2 25 g N2 1 mol N2 2 mol NO2 46.01 g NO2 35 g O2 1 mol O2 2 mol NO2 46.01 NO2 28.02 g N2 1 mol N2 1 mol NO2 32.00 g O2 2 mol O2 1 mol NO2 50.32 g NO2 is produced NO2 is the excess reactant, it will be left over after the reaction. O2 is the limiting reactant, this reactant will determine how much product is made.

Page 5 – Bottom - % Yield Done the same way if you need to figure out the amount of product use the limiting reactant steps. We will use the same balanced equation: N2 + 2O2  2NO2 We know that 50.32 g NO2 is produced. The amount of product you calculate from the limiting reactant problem is the theoretical yield. 50.32 g NO2 = theoretical yield Ex. Problem: When this experiment was performed in the lab, the chemist was able to produce 48.45 g of NO2. 48.45 g NO2 = actual yield %Yield = actual yield X 100 theoretical yield % Yield = 48.45 X 100 50.32 = 96.3 %