1. Stoichiometry: A calculation based on a balanced equation. Granada Hills Charter High School

2. Two types of Stoichiometry: Straight-forward: You are given information about one reactant and asked to calculate information about another reactant or product. Limiting Reactant: Given two quantities of two different reactants, you are asked to solve for the amount of product possible (there is a competition between reactants.)

3. Example of Straight-forward Stoichiometry: Carbon monoxide gas reacts with hydrogen gas to produce CH 3 OH gas. If you are given 88 grams of carbon monoxide, how many moles of CH 3 OH could you produce? Write and balance the equation first! Write the amounts of given and indicate the goal so that you have some direction.

4. Straight Stoichiometry: CO (g) + 2 H 2 (g)  CH 3 OH 88.0 g ? = mol Molar Mass of CO = 28.0 g/mol 88.0 g CO1 mol CO1 mol CH 3 OH= 3.10 mol CH 3 OH 28.0 g1 mol CO

5. Example of Limiting Reactant Stoichiometry: Given 88 g of CO gas and 20.0 g of H 2 gas, how many mol of CH 3 OH gas could be produced? Notice that there are TWO quantities of reactants given. You would not only complete the first calculation as detailed previously, but would continue to calculate the product possible from the 20.0 g of H 2 gas.

6. Calculating the product possible from the Second Reactant Molar Mass of H 2 = 2.016 g/mol 20.0 g H 2 1 mol H 2 1 mol CH 3 OH= 4.96 mol CH 3 OH 2.016 g2 mol H 2 From the previous calculation it was found that the CO could produce only 3.10 mol of the product. Comparing this with the amount produced by the H 2 we see that the CO is the limiting reactant and it determines the maximum amount of product possible before this reactants runs out.

7. Percent Yield Continuing with our problem: In laboratory you produced 2.8 mol of CH 3 OH. What was your percent yield? actual x 100% = % Yield theoretical 2.8 mol x 100 = 90% Yield 3.10 mol Actually produced in lab The product calculated from our previous limiting reactant problem