Stoichiometry

Molar Mass Calculations molar mass Grams Moles When performing your conversions, you want to make your diagonals the same unit so the units will cancel out. 15.0 grams Al 1 mol 26.98 grams Al

Practice Problems 1) How many moles of carbon are in 26 g of carbon? 26 g C 1 mol C 12.01 g C = 2.2 mol C 2) How many grams of carbon are in 3.45 moles of carbon? 3.45 mol C 12.01 g C 1 mol C = 41.4 grams C

Molar Mass Activity

Mole Ratios The mole ratios come from the balanced coefficients in the equation. They are required when changing from moles of one compound to moles of a different compound Example: How many moles of chlorine are needed to react with 5 moles of sodium 2 Na + 1 Cl2  2 NaCl 5 moles Na 1 mol Cl2 2 mol Na = 2.5 moles Cl2

Mole-Mass OR Mass to Mole Conversions Most of the time, the answer will be in grams. We use molar ratios to switch compounds, and use molar mass conversion to get into grams. Example: 5.00 moles of sodium will require how many grams of chlorine? 2 Na + Cl2  2 NaCl 5.00 moles Na 1 mol Cl2 70.90g Cl2 2 mol Na 1 mol Cl2 = 177g Cl2

Mass-Mass Conversions Most often we are given a starting mass and want to determine the mass of a final product . This is called the (theoretical yield) This is the most common type of conversion, but we will see others.

Mass-Mass Conversion N2 + 3 H2  2 NH3 Ex. Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen. N2 + 3 H2  2 NH3 2.00g N2 1 mol N2 2 mol NH3 17.06g NH3 28.02g N2 1 mol N2 1 mol NH3 = 2.44 g NH3

Limiting Reactant: Example 10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 2 Al + 3 Cl2  2 AlCl3 Start with Al: Now Cl2: 10.0 g Al 1 mol Al 2 mol AlCl3 133.5 g AlCl3 27.0 g Al 2 mol Al 1 mol AlCl3 = 49.4g AlCl3 35.0g Cl2 1 mol Cl2 2 mol AlCl3 133.5 g AlCl3 71.0 g Cl2 3 mol Cl2 1 mol AlCl3 = 43.9g AlCl3

Finding the Amount of Excess By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess. Can we find the amount of excess potassium in the previous problem?

Finding Excess Practice 15.0 g of potassium reacts with 15.0 g of iodine. 2 K + I2  2 KI We found that Iodine is the limiting reactant. We need to determine how many grams of Potassium (excess) was used, then subtract! 15.0 g I2 1 mol I2 2 mol K 39.1 g K 254 g I2 1 mol I2 1 mol K = 4.62 g K USED! 15.0 g K – 4.62 g K = 10.38 g K EXCESS Given amount of excess reactant Amount of excess reactant actually used Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!