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Limiting Reagent  1. Limits or determines the amount of product that can be formed  2. The reagent that is not used up is therefore the excess reagent.

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Presentation on theme: "Limiting Reagent  1. Limits or determines the amount of product that can be formed  2. The reagent that is not used up is therefore the excess reagent."— Presentation transcript:

1 Limiting Reagent  1. Limits or determines the amount of product that can be formed  2. The reagent that is not used up is therefore the excess reagent  These types of problems require 2 sets of tracks. Quantities of both reagents will be given. Therefore, you need to find out which one is the limiting reagent.  1. Limits or determines the amount of product that can be formed  2. The reagent that is not used up is therefore the excess reagent  These types of problems require 2 sets of tracks. Quantities of both reagents will be given. Therefore, you need to find out which one is the limiting reagent.

2 Limiting Reagent  One track to determine limiting reagent  A second track to determine product  One track to determine limiting reagent  A second track to determine product

3 Limiting Reagent Example problem  How many grams of copper (I) Sulfide can be produced when 80.0 grams of Cu reacts with 25.0 grams of sulfur?  2Cu + S --> Cu 2 S  Pick a reactant and calculate how much of the other reactant is needed. 80.0g Cu 1mol Cu 1mol S 32.1g S 63.5g Cu 2mol Cu 1mol S = 20.2g S So, 20.2 g of S is needed; 25.0g is supplied Plenty of S; therefore, Cu is limiting reagent. Use Cu to solve the problem 80.0g Cu 1mol Cu 1mol Cu 2 S 159.1g Cu 2 S 63.5g Cu 2mol Cu 1mol Cu 2 S = 1.00x10 2 g Cu 2 S  How many grams of copper (I) Sulfide can be produced when 80.0 grams of Cu reacts with 25.0 grams of sulfur?  2Cu + S --> Cu 2 S  Pick a reactant and calculate how much of the other reactant is needed. 80.0g Cu 1mol Cu 1mol S 32.1g S 63.5g Cu 2mol Cu 1mol S = 20.2g S So, 20.2 g of S is needed; 25.0g is supplied Plenty of S; therefore, Cu is limiting reagent. Use Cu to solve the problem 80.0g Cu 1mol Cu 1mol Cu 2 S 159.1g Cu 2 S 63.5g Cu 2mol Cu 1mol Cu 2 S = 1.00x10 2 g Cu 2 S

4 Limiting Reagent Example Problem - Your Turn  How many grams of hydrogen can be produced when 5.00g of Mg is added to 6.00 g of HCl?  Mg + 2 HCl --> MgCl 2 + H 2  Pick a reactant and calculate how much of the other reactant is needed. 5.00g Mg 1mol Mg 2mol HCl 36.5g HCl 24.3g Mg 1 mol Mg 1mol HCl = 15.0g HCl Need 15.0g HCl; have 6.00 g HCl Not enough HCl; therefore, HCl is limiting reagent Use HCl to solve the problem 6.00g HCl 1mol HCl 1mol H 2 2.0g H g HCl 2mol HCl 1mol H 2 = g H 2  How many grams of hydrogen can be produced when 5.00g of Mg is added to 6.00 g of HCl?  Mg + 2 HCl --> MgCl 2 + H 2  Pick a reactant and calculate how much of the other reactant is needed. 5.00g Mg 1mol Mg 2mol HCl 36.5g HCl 24.3g Mg 1 mol Mg 1mol HCl = 15.0g HCl Need 15.0g HCl; have 6.00 g HCl Not enough HCl; therefore, HCl is limiting reagent Use HCl to solve the problem 6.00g HCl 1mol HCl 1mol H 2 2.0g H g HCl 2mol HCl 1mol H 2 = g H 2

5 Limiting Reagent Example problem- Your Turn  Acetylene (C 2 H 2 ) will burn in the presence of oxygen. How many grams of water can be produced by the reaction of 2.40 mol of acetylene with 7.4 mol of oxygen?  2 C 2 H O 2 --> 4 CO H 2 O  Pick a reactant and calculate how much of the other reactant is needed  2.40 mol C 2 H 2 5 mol O 2  2 mol C 2 H 2 = 6.00 mol O 2  Need 6.00mol O 2 ; have 7.4mol O 2  Plenty of O 2 ; so, C 2 H 2 is L.R.  2.4mol C 2 H 2 2mol H 2 O 18.0g H 2 O  2mol C 2 H 2 1mol H 2 O2  = 43.2 g H 2 O  Acetylene (C 2 H 2 ) will burn in the presence of oxygen. How many grams of water can be produced by the reaction of 2.40 mol of acetylene with 7.4 mol of oxygen?  2 C 2 H O 2 --> 4 CO H 2 O  Pick a reactant and calculate how much of the other reactant is needed  2.40 mol C 2 H 2 5 mol O 2  2 mol C 2 H 2 = 6.00 mol O 2  Need 6.00mol O 2 ; have 7.4mol O 2  Plenty of O 2 ; so, C 2 H 2 is L.R.  2.4mol C 2 H 2 2mol H 2 O 18.0g H 2 O  2mol C 2 H 2 1mol H 2 O2  = 43.2 g H 2 O


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