# Section 6.6—Limiting Reactants What happens if you don’t add reactants in a molar ratio?

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Section 6.6—Limiting Reactants What happens if you don’t add reactants in a molar ratio?

Planning a Meal You go to the grocery store and you buy 1 package of Brats (5 Brats), 1 package of cheese (16 slices) and 1 package of hot dog buns (8 buns). If you use all of these…You can make this many… 5 Brats 5 meals 16 slices of cheese 8 hot dog buns 16 meals 8 meals So you have the possibility of making 5, 16 or 8 meals…which is it? You’ll never get the chance to make 8 or 16 meals…you’ll run out of Brats after 5. Once you run out of one component, you have to stop making meals.

What’s a limiting reactant? Limiting Reactant – The reactant that runs out and causes the reaction to stop. In the previous example, the Brats were the limiting reactant—once they were gone, you had to stop! Once even one of the reactants runs out, the reaction stops…it can’t make any more product.

Excess reactant The reactant that is not completely used up in a reaction

When solving: Both reactants will be given info. Use stoichiometry to find the (mass, moles, vol. etc) of EACH reactant Whichever number is smaller will be your limiting reactant

Limiting Reactant Example Example: How many moles of H 2 O is produced when 2.3 moles O 2 and 2.3 moles H 2 react? 2 H 2 + O 2  2 H 2 O

From balanced equation: 2 mole H 2  2 mole H 2 O 1 mole O 2  2 mole H 2 O Limiting Reactant Example 2.3 mole O 2 mole O 2 mole H 2 O = ________ mole H 2 O 1 2 4.6 2.3 mole H 2 mole H 2 mole H 2 O = ________ mole H 2 O 2 2 2.3 Example: How many moles of H 2 O is produced when 2.3 moles O 2 and 2.3 moles H 2 react? 2 H 2 + O 2  2 H 2 O

From balanced equation: 2 mole H 2  2 mole H 2 O 1 mole O 2  2 mole H 2 O Limiting Reactant Example 2.3 mole O 2 mole O 2 mole H 2 O = ________ mole H 2 O 1 2 4.6 2.3 mole H 2 mole H 2 mole H 2 O = ________ mole H 2 O 2 2 2.3 Example: How many moles of H 2 O is produced when 2.3 moles O 2 and 2.3 moles H 2 react? 2 H 2 + O 2  2 H 2 O Limiting reactant!

Let’s Practice Example: If you react 10.5 g of NaOH and 7.5 g of BaCl 2, how many grams NaCl is produced? 2 NaOH + BaCl 2  Ba(OH) 2 + 2 NaCl

From balanced equation: 2 mole NaOH  2 mole NaCl 1 mole BaCl 2  2 mole NaCl Let’s Practice 10.50 g NaOH g NaOH mole NaOH = _______ g NaCl 40.00 1 15.3 Molecular masses: 40.00 g NaOH = 1 mole NaOH 171.35 g Ba(OH) 2 = 1 mole Ba(OH) 2 58.44 g NaCl = 1 mole NaCl mole NaOH mole NaCl 2 2 g NaCl 1 58.44 Example: If you react 10.5g of NaOH and 7.5g of BaCl 2, how many grams NaCl is produced? 2 NaOH + BaCl 2  Ba(OH) 2 + 2 NaCl 7.50 g BaCl 2 g BaCl 2 mole BaCl 2 = _______ g NaCl 171.35 1 5.12 mole BaCl 2 mole NaCl 1 2 g NaCl 1 58.44

Lets Practice 2Na + Cl 2  2NaCl Suppose that 6.70 mol of Na reacts with 3.20 mol of Cl 2.  1) What is the limiting reactant?  How many moles of NaCl are produced?

Lets Practice 2Cu (s) + S (s)  Cu 2 S (s)  What is the limiting regent when 80.0g Cu reacts with 25.0g S?  How grams of Cu 2 S will be produced?