Day SOLVE: 1.5e24 molecules of SO 2 = ____ grams of SO 2 ? S 2 + 2O 2  2SO 2 And how many moles of S 2 ?

Limiting Reactant – the reactant that limits the amounts of the other reactants that can combine and the amount of product that can form in a chemical reaction – the one you will run out of Excess Reactant – any substance that is not used up completely in a reaction – left over Limiting Reactants

To determine the Limiting Reactant – use either reactant to calculate the amounts necessary for reaction Example: 4NH 3 (g) + 6NO(g) 5N 2 (g) + 6H 2 O(l) Given the following, determine the limiting reactant: 40 g NH 3 41 g NO

Limiting Reactants Example: 4NH 3 (g) + 6NO(g) 5N 2 (g) + 6H 2 O(l) Given the following, determine the limiting reactant: 40 g NH 3 41 g NO

2Na + Cl 2 → 2NaCl Limiting Reactants Start with 4 moles of Na and 3 moles of Cl 2 : limiting = ? excess = ?

2Na + Cl 2 → 2NaCl Limiting Reactants Start with 26 grams of Na and 35.5 grams of Cl 2 : limiting = ? excess = ?

Day If you start with 2.5 moles of solid aluminum and allow it to react with 2.5 moles of CuCl 2 which reactant will be left over at the end? 2.Why is today Mole Day?

Day Which reactant determines your theoretical yield?

Percent Yield = Percent Yield Actual Yield Theoretical Yield X 100  the percent yield is a measure of the efficiency of a reaction carried out in the lab. - a perfect percent yield would be a 100%

Example # 1: Calcium carbonate, which is found in seashells, is decomposed by heating. The reaction is below: CaCO 3(s)  CaO (s) + CO 2(g) What is the theoretical yield of CaO if 24.8 grams of CaCO 3 is heated? If only 12 grams of CaO are recovered, what is the percent yield? What could be the cause?

12 Quarterly Practice A solution containing 3.50 g sodium phosphate is mixed with a solution containing 6.40 g barium nitrate. After the reaction is complete, 4.85 g barium phosphate are collected. 1. What is the limiting reagent? 2. What is the theoretical yield of barium phosphate? 3. What is the percent yield of barium phosphate? 2Na 3 PO 4(aq) + 3Ba(NO 3 ) 2(aq)  6NaNO 3(aq) + Ba 3 (PO 4 ) 2(s) molar masses: Na 3 PO 4 = Ba(NO 3 ) 2 = Ba 3 (PO 4 ) 2 = 601.9

13 Quarterly Practice A solution containing 3.50 g sodium phosphate is mixed with a solution containing 6.40 g barium nitrate. After the reaction is complete, 4.85 g barium phosphate are collected. 1. What is the limiting reagent? 2. What is the theoretical yield of barium phosphate? 3. What is the percent yield of barium phosphate? 2Na 3 PO 4(aq) + 3Ba(NO 3 ) 2(aq)  6NaNO 3(aq) + Ba 3 (PO 4 ) 2(s) If you start with 3.50 g Na 3 PO 4 you NEED 8.37 g Ba(NO 3 ) 2 So Barium nitrate is limiting and you start your theoretical with 6.40 g barium nitrate Theoretical yield = 4.91 % yield = 98.8 %

Review sections 12.1 and 12.2 and complete # 8 on page 389, # 11 on page 391, # 13 on page 393, and # 23 on page 398 – due tomorrow Assignment

Get to # 11 Hand in assignment from Friday

Review sections 12.1 and 12.2 and complete #s 5, 6, 8, 9, and 10 on page 389, #s 11 and 12 on page 391, and #s 13 and 14 on page 393 – due beginning of class Tuesday Complete #s on page 398 – due Wednesday Assignment