LIMITING REAGENT Just a Quick Note Gas occupies 22.4L/mol at STP (0 o C) Gas occupies 24.5L/mol at NTP (25 o C) Limiting Reactant In a chemical reaction where arbitrary amounts of reactants are mixed and allowed to react, the one that is used up first is the limiting reactant. A portion of the other reactants remains. There is a systematic procedure for finding the limiting reagent based on the reactant ratio (RR) defined as the ratio of the number of moles of a reactant to its coefficient in a balanced chemical equation. The reagent with the smallest reactant ratio is the limiting reactant. The Limiting Reagent o In most chemical reactions, not all of the reactants are used up because they are not present in the exact proportions required of the reaction Eg. If Pb(NO 3 ) 2 + CuSO 4 PbSO 4 + Cu(NO 3 ) 2 ; you would need exactly 313g (1mol) of Pb(NO 3 ) 2 and 303g (1mol) of CuSO 4 to fully use each of the reactants o The limiting reagent is the reactant which is used up first in a chemical reaction (i.e. it being used up, stops the rest of the reactants from being able to fully react) o The limiting reagent can be determined mathematically using the Mole Balanced reaction! Defines stoichiometric ratios! Unbalanced (i.e., non-stoichiometric) mixture! Limited by syrup! For a Reaction of the Form: aA + bB = cC + dD If compounds A and B are present in the mole amounts called for in the balanced reaction, then the following equation is valid: To proceed, first calculate the reactant ratios for all of the reactants: From among these, choose (RR) min, the smallest reactant ratio. This identifies the limiting reactant. aA + bB = cC + dD 2Al(s) + 6HCl(g) 2AlCl 3 (s) + 3H 2 (g) Consider the reaction above. If we react 30.0 g Al and 20.0 g HCl, how many moles of aluminum chloride will be formed? 30.0 g Al20.0g HCl Limiting reactant = reactant with RR min = Limiting Reactant EXAMPLE MOLES = M/MM = 30 / 27 = RRa = (Moles of Al) / a = / 2 = MOLES = M/MM = 20/ ( ) = RRb = (Moles of HCl) / b = / 6 = Calculate the Moles of AlCl 3 from: Moles of Limiting Reagent x (need/have) = x 2/ 6 = Finding amount of Product from Limiting Reagent 2Al(s) + 6HCl(g) 2AlCl 3 (s) + 3H 2 (g) 1.4KO 2 + 2H 2 O 4KOH + 3O 2 : 0.15mol KO 2 and 0.10mol H 2 O a.Which is the limiting reagent? b.How many moles of O 2 can be produced? c.What is the volume of O 2 is produced at STP? 2.CO + H 2 CH 3 OH: 35.4g CO and 10.2g H 2 a.Balance the equation b.Which is the limiting reagent? c.How many grams of the excess reagent are left at the end of reaction? ( Hint [Need / have] x sentence ) use this to calculate the moles of H2 used 1.. a.KO 2 is the limiting reagent b.0.11mol of O 2 c.2.52 L 2.. a.CO + 2H 2 CH 3 OH b.CO is the limiting reagent c.5.1g of H 2 Work on Worksheet