Limiting & Excess. Cookie Stoichiometry To make one batch, it requires: 2 eggs 2 cups of flour 2/3 cups of butter How many batches could you make with…

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Presentation transcript:

Limiting & Excess

Cookie Stoichiometry To make one batch, it requires: 2 eggs 2 cups of flour 2/3 cups of butter How many batches could you make with… – 4 eggs – 6 cups of flour – 2/3 cups of butter – The cups of butter was your limiting reagent – The eggs and flour were your excess reagents – But how could you show this mathematically…?

2 Together…. 3 on your Own! Whiteboard!

Example g of H 2 react with g of O 2 according to the reaction 2 H 2 + O 2  2 H 2 O Which reactant is limiting and which reactant is in excess?

Example 1 1. Calculate the number of moles of a product formed. Mole of H 2 O (based on H 2 ) =20.00 g H 2 x 1 mol H 2 x 2 mol H 2 O = 10.0 mole H 2 O 2.0 g H 2 2 mol H 2 Mole of H 2 O (based on O 2 ) =100.0 g O 2 x 1 mol O 2 x 2 mol H 2 O = 6.25 mol H 2 O 32.0 g O 2 1 mol O 2

Example 1 Determine the limiting and excess reactants. O 2 is the limiting reactant. H 2 is the excess reactant. 2. How much H 2 (in grams) is left after O 2 runs out?

Example 1 3. To find mass of H 2 in excess, find the mass of H 2 which will react based on the mass of the limiting reactant. Mass of H 2 (based on O 2 ) =100.0 g O 2 x 1 mol O 2 x 2 mol H 2 x 2.0 g H 2 =12.5 g H g O 2 1 mol O 2 1 mol H 2

Example 1 4. Then, subtract the mass of H 2 which reacts from the starting mass of H 2. Mass of H 2 (in excess) = mass H 2 (at start) – mass H 2 (reacted) = g – 12.5g = 7.5 g Take Attendance

Example 2 If 56.8 g of FeCl 2, 14.9 g of KNO 3 and 40.0 g of HCl are mixed according to the reaction FeCl 2 + KNO 3 + HCl  FeCl 3 + NO + H 2 O + KCl a)What is the limiting reactant? b)How many grams of each “excess” reactant left in the reaction vessel?

Example 2 If 56.8 g of FeCl 2, 14.9 g of KNO 3 and 40.0 g of HCl are mixed according to the reaction 3 FeCl 2 + KNO HCl  3 FeCl 3 + NO + 2 H 2 O + KCl a)What is the limiting reactant? b)How many grams of each “excess” reactant left in the reaction vessel?

Example 2 If 56.8 g of FeCl 2, 14.9 g of KNO 3 and 40.0 g of HCl are mixed according to the reaction 3 FeCl 2 + KNO HCl  3 FeCl 3 + NO + 2 H 2 O + KCl a)What is the limiting reactant? KNO 3 b)How many grams of each “excess” reactant left in the reaction vessel?

Example 2 If 56.8 g of FeCl 2, 14.9 g of KNO 3 and 40.0 g of HCl are mixed according to the reaction 3 FeCl 2 + KNO HCl  3 FeCl 3 + NO + 2 H 2 O + KCl a)What is the limiting reactant? KNO 3 b)b) How many grams of each “excess” reactant are actually present in excess? FeCl 2 = 0.8g, HCl = 18.5g

Example 3 If 3.2 g of CuSO 4, 2.5 g of water and 3.0 g of SO 2 are reacted together in the reaction: CuSO 4 + H 2 O + SO 2  Cu + H 2 SO 4 Which reactant is the limiting reactant? What is the mass of each of the excess reactants?

Example mL of M ammonia (NH 3 ) solution is mixed with 15.0 g of solid magnesium to produce magnesium nitride and hydrogen gas. Which reactant is in excess? How much (in grams) of the excess reactant remains after the reaction is complete?

Example 5: For the reaction: TiO 2(aq) + B 4 C (g) + C (s)  TiB 2(aq) + CO (g) If 11.5 L of M TiO 2, 455 g of solid carbon and 184 L of B 4 C gas at STP react together. Find the amount (based on the original units given) of the reactants left over.

Example 5: For the reaction: 2TiO 2(aq) + B 4 C (g) + 3 C (s)  2TiB 2(aq) + 4CO (g) If 11.5 L of M TiO 2, 455 g of solid carbon and 184 L of B 4 C gas at STP react together. Find the amount (based on the original units given) of the reactants left over.