percent yield and limiting reagents - oakparkusd.org · limiting reactants • the substance that...
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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 =×