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Why do I care??? Balanced equations give exact mole ratios When the reactions are carried out, the reactants are usually not available in those same ratio amounts One reactant will be in excess, while one will be limited. The reaction will continue until the limiting reactant runs out.
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New Terms!!! Limiting reactant- limits the extent of the reactants and determines the amount of product produced Excess reactants- reactants left over after the reaction has stopped
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Here we go… Steps for finding the product when there is a LR. 1.Find # of moles of each reactant 2. Determine whether reactants are in correct mole ratio 3. Change moles of limiting reactant to moles of product 4. Change moles of product to grams of product
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Example Time!!! S 8 (l) + 4Cl 2 4S 2 Cl 2 (l) 200.0gS reacts with 100.0gCl:mass of product =? Step 1: Find # of moles of each reactant 100.0g Cl 2 X 1mol Cl 2 =1.410 mol Cl 2 70.91g Cl 2 200.0g S 8 X 1mol S 8 = 0.7797 mol S 8 256.5g S 8
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Step 2: 1.410 mol Cl 2 available * 1 mol S 8 = 0.3525 mol S 8 4 mol Cl 2 We have 0.7797 mol S 8 So Cl 2 must be the LR! Determine which reactant runs out first by converting one into the other and comparing mole amounts
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Step 3: 1.410 mol Cl 2 X 4 mol S 2 Cl 2 = 1.410 mol S 2 Cl 2 4 mol Cl 2 Step 4: Change moles of product to grams of product 1.410 mol S 2 Cl 2 X 135.0g S 2 Cl 2 = 190.4g S 2 Cl 2 1 mol S 2 Cl 2 Change moles of limiting reactant to moles of product
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Now we are ready to find out how much excess reactant is left over! Step 5: Change moles of limiting reactant to moles of excess reactant. 1.410 mol Cl 2 X 1 mol S 8 = 0.3525 mol S 8 4 mol Cl 2 Step 6: Change actual used moles of excess reactant to grams. 0.3525 mol S 8 X 256.5g S 8 = 90.42g S 8 1 mol S 8
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Almost Done!!! Step 7: Subtract amount used from amount available. 200.0g S 8 available – 90.42g S 8 needed = 109.6g S 8 excess AND YOU ARE DONE!!!
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Almost… Now you have to try it without me! As a group: Complete Practice Problems Page 368 #20 &21 On your own: Complete Page 878 Section 12-3 #11 & 12
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