Percent Yield
and Limiting Reagents
February 17, 2011
Chapter 12
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Limiting Reactants: Identifying and Calculating the Amount of Product, Study notes of Stoichiometry
The concept of limiting reactants and excess reactants in chemical reactions. It includes examples and steps for solving problems involving the identification of limiting reactants and the calculation of the amount of product produced. The document also covers mole ratios and the conversion of moles to grams.
What you will learn
- What is the relationship between moles and grams in chemical reactions?
- What is the role of excess reactants in a chemical reaction?
- What is a limiting reactant in a chemical reaction?
- How do you identify the limiting reactant in a given reaction?
- How do you calculate the amount of product produced in a reaction using the limiting reactant?
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Percent Yield
and Limiting Reagents
February 17, 2011 Chapter 12
LIMITING REACTANTS
February 19, 2015
Something to Think About
- Given: 3 meat patties, 4 hamburger buns, 6 cheese slices, and 7 tomato slices
- Question: How many sandwiches can be made if each sandwich MUST have: 1 bun, 1 meat patty, 1 cheese slice, and 2 slices of tomato?
Cheeseburger Problem
- What ingredient(s) are left over (in excess)? How much of each is left? - 1 bun, 3 cheese slices, 1 tomato
- What ingredient(s) are used completely (limiting)?
- Meat patties
Example Problem #
- A 1.4g sample of magnesium is treated
with 8.1g of hydrochloric acid to
produce magnesium chloride and
hydrogen gas. How many grams of
hydrogen are produced?
__ **Mg (s)
- __ HCl (aq)** **__ MgCl 2 (aq)
- __ H 2 (g)**
Example Problem #
- Balance the equation Mg (s) + 2 HCl (aq) MgCl 2 (aq) + H 2 (g) 1.4 g 8.1 g x g?
- Change masses of reactants in moles Mg: 1.4 g_ = 0.0576 Moles 24.3 g HCl: 8.1 g_ = 0.222 Moles 36.5 g Limiting Reactant Excess Reactant
Example Problem #
- The reaction between solid white phosphorus (P 4 ) and oxygen (O 2 ) produces solid tetraphosphorus decoxide (P 4
O
10
Determine the mass of P 4
O
10 formed if 25.0 g of phosphorus and 50.0 g of oxygen are combined.
- What is the limiting reactant?
- What is the excess reactant?
- How much of the excess reactant remains after the reaction?
Let’s Solve the Problem
P
4
+ 5O
2
→ P
4
O
10
- Calculate the number of moles available of each reactant.
0.202 Moles
123.9 g P
25.0g P
P :
4 4 4
1.56 Moles
31.998 g O
50.0g O
O :
2 2 2
25.0 g 50.0 g
Calculate Amount of Product P 4
+ 5O
2
→ P
4
O
10
- Set Up Mole Ratio and Solve! 0.202 M x g? 0.202 Moles 0.202 Moles P x Moles P O 1 1 P P O 4 4 10 4 4 10 = = = 4 10 4 10
0.202 Moles × 283.9 g P O = 57.3g P O
Determine Excess Remaining
- Since Oxygen is in excess, only part of the available O 2 is consumed
- Use the limiting reactant (.202 M) to determine the mass of the oxygen consumed 2 4 2 4 2
1.01Moles O
.202 Moles P
x Moles O
x
P
O
1.01 MolesO 32. 0 g M 32. 0 g 2 × =