Percent Yield and Limiting Reagents

February 17, 2011 Chapter 12

LIMITING REACTANTS February 19, 2015

Limiting Reactants •  The substance that is used up first is the

limiting reactant

•  The substance that is left over after the reaction is the excess reactant

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

Steps for Solving Limiting Reagent Problems Limiting Reactant Problems involve 2 steps: 1. Identify the Limiting Reactant (LR)

•  Calculate the number of moles obtained from each reactant in turn

•  The reactant that gives the smaller amount of product is the Limiting Reactant

2. Calculate the amount of product obtained from the Limiting Reactant •  Set up a mole ratio to solve the problem

Example Problem #1 •  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) __ MgCl2(aq) + __ H2(g)

Example Problem #1 •  Balance the equation

Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(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

Determine Amount of Hydrogen Produced (Mole-Mass Problem)

Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g) 0.0576M x g?

•  Set up Mole Ratio using Limiting Reactant (Mg) Mg = 1 Mole = 0.0576 Moles = 0.0576 Moles H2 1 Mole x Moles •  Convert Moles of H2 to grams 0.0576 Moles H2 × 2.02 gram H 2 = 0.12 g H 2

Example Problem #2 •  The reaction between solid white phosphorus

(P4) and oxygen (O2) produces solid tetraphosphorus decoxide (P4 O10). Determine the mass of P4O10 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 P4 + 5O2 → P4O10

•  Calculate the number of moles available of

each reactant. Moles 0.202

P g 123.9P g 25.0:P

4

44 =

Moles 1.56 O g 31.998

O 50.0g :O2

22 =

25.0 g 50.0 g

Determine Limiting Reactant and Excess Reactant •  Phosphorous is the Limiting Reactant •  Oxygen is the Excess Reactant •  Use the moles of limiting reactant

(Phosphorous) to calculate the grams of tetraphosphorus decaoxide that can be produced

Calculate Amount of Product P4 + 5O2 → P4O10

•  Set Up Mole Ratio and Solve!

0.202 M x g?

Moles 0.202 P Moles 0.202

OP Molesx 11

POP

4

104

4

104 ===

104104 OP g 57.3 OP g 283.9 Moles 0.202 =×

Determine Excess Remaining •  Since Oxygen is in excess, only part of the

available O2 is consumed •  Use the limiting reactant (.202 M) to

determine the mass of the oxygen consumed

24

2

4

2 O Moles 1.01 P Moles .202

O Molesx x 15

PO ==

gMg 0.320.32O Moles 1.01 2 =×

Determine Excess Remaining •  Subtract the mass of oxygen needed from

the amount of oxygen available to calculate the amount of oxygen in excess

O2(available) - O2(needed) = O2 (excess)

50.0 g O2- 32.3 g O2 = 17.7 g O2