Download presentation
Presentation is loading. Please wait.
Published byLayton Miers Modified over 10 years ago
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 2 36.5g HCl 2mol HCl 1mol H 2 = 0.164 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 2 36.5g HCl 2mol HCl 1mol H 2 = 0.164 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 2 + 5 O 2 --> 4 CO 2 + 2 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 2 + 5 O 2 --> 4 CO 2 + 2 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
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.