# Limiting & Excess Reagents Chemistry 11 March 19, 2010.

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Limiting & Excess Reagents Chemistry 11 March 19, 2010

So what happens if you react…

…THIS MUCH SULPHUR…

…with this much hydrogen?

Limiting & Excess Reactants We learned that reactants in a chemical reaction react in very specific ratios (mole ratios) 2 H 2 + O 2  2 H 2 O In order to react the reactants perfectly, you need to measure the exact masses (moles) of both reactants perfectly (“down to the molecule”) HOWEVER, in reality, you cannot measure out the exact masses of all reactants! One reactant will always be in excess, the other reactant will be called “limiting”

Limiting & Excess Reactants Chemicals continue to react until one reactant is completely used up (the reaction then stops) The “excess reactant” is the reactant that is still left after the reaction has completed to form products

When Is Enough Enough? How do you know if there is too much of one reactant? You must know the mole ratio between the reactants 2 H 2 + O 2  2 H 2 O We know that 4 mol of H 2 will react with 2 mol O 2 What if you had 4 mol H 2 and 5 mol O 2 ? Only 2 mol of O 2 would react 3 mol O 2 left over

Limiting & Excess Reactants Consider the reaction of hydrogen with sulphur to yield hydrogen sulphide gas: 8 H 2 + S 8  8 H 2 S If you have 6 moles of H 2 gas and 7 moles of S 8. Which reactant is in excess? Which reactant is limiting? How many moles of water will form?

Calculations with Limiting & Excess Reactants 8 H 2 + S 8  8 H 2 S TWO WAYS to look at this problem: “Hydrogen perspective”: (6 mol H 2 ) × (1/8) = 0.75 mol S 8 But we have 7 mol of S 8 ! Therefore, there is an excess of S 8 (automatically, H 2 is limiting) “Sulphur perspective”: (7 mol of S 8 ) × (8/1) = 56 mol H 2 But we only have 5 mol H 2 ! Therefore, H 2 is “limiting” (automatically, S 8 is excess)

Calculations with Limiting & Excess Reactants 8 H 2 + S 8  8 H 2 S So how many moles of H 2 S are formed? You must figure out first which reactant is limiting. We found that H 2 was limiting (6 mol H 2 ) Therefore, 6 mol H 2 will react with 0.75 mol S 8 to yield 6 mol H 2 S There will be 6.25 mol S 8 left over

Before, During & After the Reaction 8 H 2 + S 8  8 H 2 S Stoich Coeff. SubstInitial moles Moles reacted Final moles 8H 2(g) 6 mol 0 mol 1S 8(s) 7 mol0.75 mol6.25 mol 8H 2 S (g) 0 mol--6 mol

Moles Bridge Basics Moles of chemical 1 Moles of chemical 2 # of particles of chemical 1 (atoms/molecules) Mass (g) of chemical 1 # of particles of chemical 2 (atoms/molecules) Mass (g) of chemical 2 g/mol stoichiometric coefficients Volume of Chemical 2 (22.4 L/mol) Volume of Chemical 1 (22.4 L/mol) 6.02 x 10 23 molecules/mol 6.02 x 10 23 molecules/mol