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STOICHIOMETRY TUTORIAL

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(1-2-3) General Approach For Problem Solving:

1. Clearly identify the Goal or Goals and the UNITS involved. (What are you trying to do?)

2. Determine what is given and the UNITS.

3. Use conversion factors (which are really ratios) and their UNITS to CONVERT what is given into what is desired.

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Sample problem for general problem solving.Sam has entered into a 10 mile marathon. Use ALL of the following conversions (ratios) to determine how many inches there are in the race.5280 ft = 1 mile; 12 inches = 1 ft

1. What is the goal and what units are needed?

Goal = ______ inches2. What is given and its units?

10 miles

3. Convert using factors (ratios).

10 miles = inches633600

Units match

Given GoalConvert

Menu

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Sample problem #2 on problem solving.

A car is traveling at a speed of 45 miles per hr (45 miles/hr). Determine its speed in kilometers per second using the following conversion factors (ratios). 1 mile = 5280 ft; 1 ft = 12 in; 1 inch = 2.54 cm; km = 1 x 103; cm = 1 x 10-2m; 1 hr =60 min; 1 min = 60 s

= km s

0.020

Units Match!

Goal

c cancels cm remains

Given

This is the

same as putting k over k

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Converting grams to moles.

Determine how many moles there are in 5.17 grams of Fe(C5H5)2.

Goal

= moles Fe(C5H5)2

Given

5.17 g Fe(C5H5)2

Use the molar mass to convert grams to

moles.Fe(C5H5)2

2 x 5 x 1.001 = 10.012 x 5 x 12.011 = 120.11

1 x 55.85 = 55.85

0.0278

units match

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Stoichiometry (more working with ratios)

Ratios are found within a chemical equation.

2HCl + Ba(OH)2 → 2H2O + BaCl2 1 1

2 moles of HCl react with 1 mole of Ba(OH)2 to form 2 moles of H2O and 1 mole of BaCl2

coefficients give MOLAR RATIOS

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When N2O5 is heated, it decomposes:

2N2O5(g) → 4NO2(g) + O2(g)

a. How many moles of NO2 can be produced from 4.3 moles of N2O5?

= moles NO2

4.3 mol N2O5 8.6

b. How many moles of O2 can be produced from 4.3 moles of N2O5?

= mole O24.3 mol N2O5 2.2

2N2O5(g) → 4NO2(g) + O2(g)4.3 mol ? mol

2N2O5(g) → 4NO2(g) + O2(g)4.3 mol ? mol

Mole – Mole Conversions

Units match

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When N2O5 is heated, it decomposes:2N2O5(g) → 4NO2(g) + O2(g)

a. How many moles of N2O5 were used if 210g of NO2 were produced?

= moles N2O5210 g NO2 2.28

b. How many grams of N2O5 are needed to produce 75.0 grams of O2?

= grams N2O575.0 g O2

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gram ↔ mole and gram ↔ gram conversions

2N2O5(g) → 4NO2(g) + O2(g)210g? moles

2N2O5(g) → 4NO2(g) + O2(g)75.0 g? grams

Units match

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Aluminum is an active metal that when placed in hydrochloric acid produces hydrogen gas and aluminum chloride. How many grams of aluminum chloride can be produced when 3.45 grams of aluminum are reacted with an excess of hydrochloric acid?

First write a balanced equation.

Al(s) + HCl(aq) AlCl3(aq) + H2(g)2 6 2 3

Gram to Gram Conversions

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Al(s) + HCl(aq) AlCl3(aq) + H2(g)2 6 2 3

Now let’s get organized. Write the information below the substances.

3.45 g ? grams

Gram to Gram Conversions

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Al(s) + HCl(aq) AlCl3(aq) + H2(g)2 6 2 33.45 g ? grams

Let’s work the problem.

= g AlCl33.45 g Al

We must always convert to moles.

Now use the molar ratio.Now use the molar mass to convert to grams.

17.0

Units match

gram to gram conversions

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Molarity

Molarity is a term used to express concentration. The units of molarity are moles per liter (It is abbreviated as a capital M)

When working problems, it is a good idea to change M

into its units.

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A solution is prepared by dissolving 3.73 grams of AlCl3 in water to form 200.0 mL solution. A 10.0 mL portion of the solution is then used to prepare 100.0 mL of solution. Determine the molarity of the final solution.

What type of problem(s) is

this?

Molarity followed by

dilution.

Solutions

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A solution is prepared by dissolving 3.73 grams of AlCl3 in water to form 200.0 mL solution. A 10.0 mL portion of the solution is then used to prepare 100.0 mL of solution. Determine the molarity of the final solution.

1st:= mol L

3.73 g

200.0 x 10-3 L0.140

2nd: M1V1 = M2V2

(0.140 M)(10.0 mL) = (? M)(100.0 mL)0.0140 M = M2

molar mass of AlCl3

dilution formula

final concentration

Solutions

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50.0 mL of 6.0 M H2SO4 (battery acid) were spilled and solid NaHCO3 (baking soda) is to be used to neutralize the acid. How many grams of NaHCO3 must be used?H2SO4(aq) + 2NaHCO3 → 2H2O(l) + Na2SO4(aq) + 2CO2(g)

Solution Stoichiometry

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50.0 mL6.0 M

? g

Look! A conversion factor!



=

Our Goal

20

50.0 mL6.0 M

? g



=

Our Goal

= g NaHCO3

H2SO4

50.0 mL1 molH2SO4

NaHCO32 mol

NaHCO3

84.0 gmolNaHCO3

50.4

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Solution Stoichiometry:

Determine how many mL of 0.102 M NaOH solution are needed to neutralize 35.0 mL of 0.125 M H2SO4 solution.

First write a balancedEquation.

____NaOH + ____H2SO4 ____H2O + ____Na2SO4 2 1 2 1

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Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H2SO4 solution.

Now, let’s get organized. Place numerical Information and

accompanying UNITS below each compound.

____NaOH + ____H2SO4 ____H2O + ____Na2SO4 2 1 2 1

0.102 M

? mL

35.0 mL

Since 1 L = 1000 mL, we can use this to save on the number of conversions

Our Goal

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Determine how many mL of 0.102 M NaOH solution is needed to neutralize 35.0 mL of 0.125 M H2SO4 solution.

Now let’s get to work converting.

____NaOH + ____H2SO4 ____H2O + ____Na2SO4 2 1 2 1

0.102 M

? mL

35.0 mL

= mL NaOH

H2SO435.0 mL

H2SO40.125 mol 1000 mL H2SO4

NaOH2 mol1 mol H2SO4

1000 mL NaOH0.102 mol NaOH

85.8

Units Match


shortcut

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What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH)2?

1st write out a balanced chemical

equation


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What volume of 0.40 M HCl solution is needed to completely neutralize 47.1 mL of 0.75 M Ba(OH)2?

2HCl(aq) + Ba(OH)2(aq) → 2H2O(l) + BaCl2

0.40 M 47.1 mL0.75 M? mL

= mL HCl

Ba(OH)247.1 mL

1 mol Ba(OH)2

HCl2 mol

0.40 mol HCl

HCl1000 mL 176

Units match


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Solution Stochiometry Problem:

A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)?

First write a balanced chemical reaction.

____HCl(aq) + ____Ba(OH)2(aq) ____H2O(l) + ____BaCl2(aq)2 1 2 1 23.28 mL

0.135 mol L

25.00 mL

? mol L

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A chemist performed a titration to standardize a barium hydroxide solution. If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of the barium hydroxide solution, what was the concentration of the barium hydroxide solution in moles per liter (M)?

____HCl(aq) + ____Ba(OH)2(aq) ____H2O(l) + ____BaCl2(aq)2 1 2 1 23.28 mL

0.135 mol L

25.00 mL

? mol L

= mol Ba(OH)2 L Ba(OH)2

25.00 x 10-3 L Ba(OH)2

Units Already Match on Bottom!

0.0629

Units match on top!

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48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of 0.385 M HNO3. Determine the molarity of the Ca(OH)2 solution.

We must first write a balanced equation.


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48.0 mL of Ca(OH)2 solution was titrated with 19.2 mL of 0.385 M HNO3. Determine the molarity of the Ca(OH)2 solution.

Ca(OH)2(aq) + HNO3(aq) → H2O(l) + Ca(NO3)2(aq)2 248.0 mL 19.2 mL

0.385 M

= mol(Ca(OH)2)

L (Ca(OH)2)

19.2 mLHNO3

48.0 x 10-3L

? M

units match!

0.0770


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Limiting/Excess/ Reactant and Theoretical Yield Problems :

Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.

4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)

a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?b. Determine the limiting reactant.

4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)

First copy down the the BALANCED

equation!

Now place numerical the

information below the compounds.

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4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)


4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g) 0.15 mol 0.10 mol ? moles

Two starting amounts?

Where do we start?

Hide

one

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4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)


4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g) 0.15 mol 0.10 mol ? molesHide

Based on:KO2 = mol O2

0.15 mol KO2 0.1125

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4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)


4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g) 0.15 mol 0.10 mol ? moles


0.15 mol KO2 0.1125

Hide

Based on: H2O

= mol O20.10 mol H2O 0.150


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Limiting/Excess/ Reactant and Theoretical Yield Problems :Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.

4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g) a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?

Determine the limiting reactant.4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g)

0.15 mol 0.10 mol ? moles


0.15 mol KO2 0.1125

Based on: H2O

= mol O20.10 mol H2O 0.150

What is the theoretical yield? Hint: Which is the smallest

amount? The is based upon the limiting reactant?

It was limited by theamount of KO2.

H2O = excess (XS) reactant!

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Theoretical yield vs. Actual yield

Suppose the theoretical yield for an experiment was calculated to be 19.5 grams, and the experiment was performed, but only 12.3 grams of product were recovered. Determine the % yield.

Theoretical yield = 19.5 g based on limiting reactantActual yield = 12.3 g experimentally recovered

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4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g) If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced?

4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g) 120.0 g 47.0 g ? gHide one

Based on:KO2

= g O2 120.0 g KO2 40.51

Limiting/Excess Reactant Problem with % Yield

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4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g) If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced?

4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g) 120.0 g 47.0 g ? g

Based on:KO2

= g O2 120.0 g KO2 40.51

Based on:H2O

= g O2

Question if only 35.2 g of O2 were recovered, what was the percent yield?

Hide

47.0 g H2O 125.3

Limiting/Excess Reactant Problem with % Yield

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If a reaction vessel contains 120.0 g of KO2 and 47.0 g of H2O, how many grams of O2 can be produced?

4KO2(s) + 2H2O(l) → 4KOH(s) + 3O2(g) 120.0 g 47.0 g ? g

Based on:KO2

= g O2 120.0 g KO2 40.51

Based on:H2O

= g O247.0 g H2O 125.3

Determine how many grams of Water were left over.The Difference between the above amounts is directly RELATED to the XS H2O.

125.3 - 40.51 = 84.79 g of O2 that could have been formed from the XS water.

= g XS H2O84.79 g O2 31.83

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Calculate the molarity of a solution prepared by dissolving 25.6 grams of Al(NO3)3 in 455 mL of solution.

After you have worked the

problem, click here to see

setup answer

Try this problem (then check your answer):

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