Unformatted text preview: SOLUTION STOICHIOMETRY
Section 5.9 Zinc reacts with
acids to produce
H2 gas. If you
have 10.0 g of Zn,
what volume of
2.50 M HCl is
needed to convert
the Zn
completely? Zinc reacts with acids to produce H22gas. If you
Zinc reacts with acids to produce H gas. If you
have 10.0 g of Zn, what volume of 2.50 M HCl is
have 10.0 g of Zn, what volume of 2.50 M HCl is
needed to convert the Zn completely?
needed to convert the Zn completely? Step 3: Use the stoichiometric factor
Step
Use
0.153 mol Zn • 2 mol HCl
= 0.306 mol HCl
1 mol Zn GENERAL PLAN FOR
GENERAL PLAN FOR
STOICHIOMETRY CALCULATIONS
STOICHIOMETRY CALCULATIONS
Mass
HCl Mass
zinc Moles
zinc Stoichiometric
factor Moles
HCl Zinc reacts with acids to produce H22gas. If you
Zinc reacts with acids to produce H gas. If you
have 10.0 g of Zn, what volume of 2.50 M HCl is
have 10.0 g of Zn, what volume of 2.50 M HCl is
needed to convert the Zn completely?
needed to convert the Zn completely? Step 1: Write the balanced equation
Step
Write
Zn(s) + 2 HCl(aq) > ZnCl2(aq) + H2(g)
Step 2: Calculate moles of Zn
Step
Zn
10.0 g Zn • Concentation
HCl ACIDBASE REACTIONS
ACIDBASE REACTIONS
Titrations
Titrations
H2C2O4(aq) + 2 NaOH(aq) >
acid
base
acid
base
Na2C2O4(aq) + 2 H2O(liq)
Carry out this reaction using a TITRATION .
Carry
TITRATION Step 4: Calculate volume of HCl req’d
Step
Calculate
1.00 L
0.306 mol HCl •
= 0.122 L HCl
2.50 mol Oxalic acid, H2C2O4 Page 1 1.00 mol Zn
= 0.153 mol Zn
65.39 g Zn Step 3: Use the stoichiometric factor
Step
Use Titration
Titration
setup
setup Buret contains a
solution whose
concentration is
known exactly. Solution of
unknown
concentration Titration
Titration 1. Add solution from the
buret.
2. Reagent (base) reacts
with compound (acid) in
solution in the flask.
3. Indicator shows when
exact stoichiometric
reaction has occurred.
4. Net ionic equation
H+ + OH > H2O
5. At equivalence point
moles H + = moles OH  LAB PROBLEM #1: Standardize a
LAB PROBLEM #1: Standardize a
solution of NaOH — ii.e., accurately
solution of NaOH — .e., accurately
determine its concentration.
determine its concentration. Step 1: Calculate moles of H 2C2O4
Step
Calculate
1.065 g of H 2C2O4
(oxalic acid) requires
35.62 mL of NaOH for
titration to an
equivalence point.
What is the concentration of the NaOH? 1.065 g of H22C2O4((oxalic acid) requires 35.62
1.065 g of H C2O4 oxalic acid) requires 35.62
mL of NaOH ffortitration to an equivalence
mL of NaOH or titration to an equivalence
point. What is the concentration of the NaOH?
point. What is the concentration of the NaOH? 1.065 g of H22C2O4((oxalic acid) requires 35.62
1.065 g of H C2O4 oxalic acid) requires 35.62
mL of NaOH ffortitration to an equivalence
mL of NaOH or titration to an equivalence
point. What is the concentration of the NaOH?
point. What is the concentration of the NaOH? Step 1: Calculate moles of H 2C2O4
Step
Calculate Step 1: Calculate moles of H2C2O4
Step
Calculate = 0.0118 mol acid
Step 2: Calculate moles of NaOH req’d
Step
Calculate
= 0.0236 mol NaOH
Step 3: Calculate concentration of NaOH
Step
Calculate 1.065 g of H22C2O4((oxalic acid) requires 35.62
1.065 g of H C2O4 oxalic acid) requires 35.62
mL of NaOH ffortitration to an equivalence
mL of NaOH or titration to an equivalence
point. What is the concentration of the NaOH?
point. What is the concentration of the NaOH? = 0.0118 mol acid
Step 2: Calculate moles of NaOH req’d
Step
Calculate
= 0.0236 mol NaOH
Step 3: Calculate concentration of NaOH
Step
NaOH 0.0236 mol NaOH
= 0.663 M
0.03562 L [NaOH] = 0.663 M Page 2 1.065 g • 1 mol
= 0.0118 mol
90.04 g Step 2: Calculate moles of NaOH req’d
Step
Calculate
0.0118 mol acid • 2 mol NaOH
= 0.0236 mol NaOH
1 mol acid LAB PROBLEM #2:
LAB PROBLEM #2:
Use standardized NaOH to determine
Use standardized NaOH to
the amount of an acid in an unknown.
the amount of an acid in an unknown.
Apples contain malic acid, C4H6O5.
C4H6O5(aq) + 2 NaOH(aq) >
Na2C4H4O5(aq) + 2 H2O(liq)
76.80 g of apple requires 34.56 mL of
0.663 M NaOH for titration. What is
weight % of malic acid? 76.80 g of apple requires 34.56 mL of
76.80 g of apple requires 34.56 mL of
0.663 M NaOH ffor titration. What is
0.663 M NaOH or titration. What is
weight % of malic acid?
weight % of malic acid? 76.80 g of apple requires 34.56 mL of
76.80 g of apple requires 34.56 mL of
0.663 M NaOH ffor titration. What is
0.663 M NaOH or titration. What is
weight % of malic acid?
weight % of malic acid? 76.80 g of apple requires 34.56 mL of
76.80 g of apple requires 34.56 mL of
0.663 M NaOH ffor titration. What is
0.663 M NaOH or titration. What is
weight % of malic acid?
weight % of malic acid? Step 1:
Step
M•V =
=
Step 2:
Step Step 1:
Step
=
Step 2:
Step
= Step 1:
Step
=
Step 2:
Step
=
Step 3:
Step
= Calculate quantity of NaOH used.
Calculate
(0.663 M)(0.03456 L)
0.0229 mol NaOH
Calculate quantity of acid titrated.
Calculate 0.0229 mol NaOH • 1 mol acid
2 mol NaOH = 0.0115 mol acid Calculate moles of NaOH used.
Calculate
0.0229 mol NaOH
Calculate moles of acid titrated
Calculate
0.0115 mol acid Step 3: Calculate mass of acid titrated.
Step
Calculate
0.0115 mol acid • 134 g
= 1 .54 g
mol Calculate moles of NaOH used.
Calculate
0.0229 mol NaOH
Calculate moles of acid titrated
Calculate
0.0115 mol acid
Calculate mass of acid titrated.
Calculate
1.54 g acid Step 4: Calculate % malic acid.
Step
Calculate
1.54 g
• 1 00% = 2.01%
76.80 g Page 3 ...
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This note was uploaded on 10/19/2011 for the course CHM 2210 taught by Professor Staff during the Spring '08 term at University of Central Florida.
 Spring '08
 Staff
 Chemistry, Stoichiometry

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