1. Limiting and Excess Reactant Limiting Reactant - the reactant in a chemical reaction that limits the amount of product that can be formed. The reaction will stop when all of the limiting reactant is consumed Excess Reactant - the reactant in a chemical reaction that remains when a reaction stops when the limiting reactant is completely consumed. The excess reactant remains because there is nothing with which it can react

2. Limiting/Excess Reactant No matter how many tires there are, if there are only 8 car bodies, then only 8 cars can be made Likewise with chemistry - if there is only a certain amount of one reactant available for a reaction, the reaction must stop when that reactant is consumed whether or not the other reactant has been used up

3. The idea of excess In a balanced equation like the one below, it is often assumed that all of the reacting chemicals change into products: CaCO 3 + 2HCl →CaCl 2 + CO 2 + H2OH2O In that case, at the end of the reaction, no CaCO 3 or HCl will be left behind However, if there is a shortage of, say, CaCO 3 then the reaction will stop when the CaCO 3 runs out Some HCl will be left over, unable to react, as there is no more CaCO 3 The HCl is said to be in excess

4. Summary Chemical reaction equations give the ideal stoichiometric relationship among reactants and products However, the reactants for a reaction in an experiment are not necessarily a stoichiometric mixture In a chemical reaction, reactants that are not used up when the reaction is finished are called excess reagents Reagent that is completely used up is called the limiting reagent, because its quantity limit the amount of products formed

5. Why is this important? In lab or industry, we want to predict amount of product we can expect from a reaction Have to use stoichiometry : quantitative study of reactants and products in a reaction (mole to mole ratio) When amounts of two reactants are given, we have to solve a limiting-reactant problem

6. Mole Ratio The mole ratio is the stoichiometric ratio of reactants and products and is the ratio of the coefficients for reactants and products found in the balanced chemical equation in the reaction: 2 Mg (s) + O 2(g) → 2MgO (s) the mole ratio of: Mg : O 2 : MgO is:is: 2 : 1 : 2 That is, the complete reaction requires twice as many moles of magnesium as there are moles of oxygen

7. Example 1. A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen. Which is the limiting reactant, how much product is produced and how much excess reactant remains after the reaction has stopped? First, we need to create a balanced equation for the reaction: 4 NH 3(g) + 5 O 2(g) → 4 NO (g) + 6 H 2 O (g) 2 g 4 g mx mx g 68 g 160 g 120 g Next we can calculate how much product is produced by each reactant: mx mx (by NH 3 ) = (2 x 120) / 68 = 3.53 g mx mx (by O 2 ) = (4 (4 x 120) / 160 = 3.00 g The reactant that produces the lesser amount of product in this case is oxygen, which is thus the "limiting reactant"

8. Example 1. 4 NH 3(g) + 5 O 2(g) → 4 NO (g) + 6 H 2 O (g) 2 g 4 g mx mx g 68 g 160 g 120 g m(NH 3 ) = (4 (4 x 68) / 160 = 1.70 g NH 3 Next, to find the amount of excess reactant, we must calculate how much of the non-limiting reactant (ammonia) actually did react with the limiting reactant (oxygen): 1.70 g is the amount of ammonia that reacted, not what is left over To find the amount of excess reactant remaining, we have to subtract the amount that reacted from the amount in the original sample: 2.00 g (original sample) – 1.70 g (reacted) = 0.30 g NH 3 remaining x g

9. Stoichiometry of Precipitation Reactions To calculate the volume of the given solution that would be required to completely precipitate the given ions from the solution If you were to add salt (sodium chloride) to a solution of silver nitrate, the MOLECULAR reaction equation would be: NaCl (aq) + AgNO 3 (aq) ↔ NaNO 3 (aq) + AgCl (s) Use solubility Table to predict which compound will precipitate: AgCl is insoluble in water we know that when NaCl and AgNO 3 dissociate in water, they will yield the following ions: To fully precipitate all of the silver(I), we need a 1:1 mole ratio of silver(I):chloride. Na + (aq) + Cl − (aq) + Ag + (aq) + NO 3 − (aq)

10. Stoichiometry of Precipitation Reactions MOLECULAR reaction equation: NaCl (aq) + AgNO 3 (aq) ↔ NaNO 3 (aq) + AgCl (s) While the NET ionic equation would be: Na + (aq) + Cl - (aq) + Ag + (aq) + NO 3 - (aq) ↔ Na + (aq) + NO 3 - (aq) + AgCl (s) Then, write the OVERALL (complete) ionic equation: Ag + (aq) + Cl - (aq) ↔ AgCl (s) since sodium and nitrate are spectator ions to fully precipitate all of the silver(I), we need a 1:1 mole ratio of silver:chloride

11. Example 2. What volume of 0.05M Na 3 PO 4 is required to precipitate all the silver ions from 75ml of a 0.1M solution of AgNO 3. Calculate the mass of the precipitate formed 1. Determine the number of moles in the 0.1 Molar AgNO 3 (acts as limiting reactant): Na 3 PO 4 + 3AgNO 3 → 3NaNO 3 + Ag 3 PO 4 ↓ x moles 0.0075 moles 0.0025 moles 1 mole 3 moles 1 mole x = 0.0025 moles Na 3 PO 4 n(AgNO 3 ) = molarity x L of solution = 0.1 x 0.075 = 0.0075 moles 2. Write the balanced equation for the reaction: 3. Calculate the moles (or mmol) of the second reactant (Na 3 PO 4 )

12. Example 2. 1000 mL – 0.05 moles Na 3 PO 4 x mL – 0.0025 moles Na 3 PO 4 4. Now plug this molar amount of Na 3 PO 4 into the molarity ratio using the molarity of the solution given in the problem to determine the volume needed: V = 50 mL 0.05M Na 3 PO 4 5. Convert 0.0025 moles of precipitated silver phosphate into grams or other units, as required: m = 0.0025 moles x 419 g/mole = 1.05 g Ag 3 PO 4 Na 3 PO 4 + 3AgNO 3 → 3NaNO 3 + Ag 3 PO 4 x moles 0.0075 moles 0.0025 moles 1 mole 3 moles 1 mole

13. Kiek ml 8% H 2 SO 4 tirpalo, kurio tankis  = 1,055 g/cm 3 reikės, norint nusodinti visus bario jonus, esančius 10 g bario chlorido? Example 3. BaCl 2 + H 2 SO 4  BaSO 4  + 2HCl 10 g x 208 g 98 g Apskaičiuojame bario chlorido molių skaičių: Reakcijoje dalyvaus tiek pat molių H 2 SO 4. Apskaičiuojame, kiek gramų grynos sieros rūgšties dalyvaus reakcijoje:

14. Apskaičiuojame, kiek reikės gramų 8% sieros rūgšties tirpalo: 8 g H 2 SO 4 – 100 g tirpalo 4,7 g H 2 SO 4­ – x Reikiamą 8% sieros rūgšties tirpalo tūrį apskaičiuojame tirpalo masę padalindami iš tankio: Atsakymas. Norint nusodinti visus bario jonus, esančius 10 g BaCl 2, reikės paimti 55,7 ml 8% H 2 SO 4 tirpalo Example 3.