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Stoichiometry A Survival Guide for the Intrepid Chemistry Student Stoichiometry A Survival Guide for the Intrepid Chemistry Student

Stoichiometry A Survival Guide for the Intrepid Chemistry Student - PowerPoint Presentation

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Stoichiometry A Survival Guide for the Intrepid Chemistry Student - PPT Presentation

What is Stoichiometry The study of the mass relationships in chemical reactions Stoichiometry can be used to find the masses of compounds and elements based on the ratios found in chemical equations ID: 919415

sulfate silver wanted grams silver sulfate grams wanted stoichiometric solve mol problems steps molar produced reaction nitrate excess magnesium

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Slide1

Stoichiometry

A Survival Guide for the Intrepid Chemistry Student

Slide2

What is Stoichiometry?

The study of the mass relationships in chemical reactions.

Stoichiometry can be used to find the masses of compounds and elements based on the ratios found in chemical equations.

Stoichiometric calculations are based on the relationship between mass and moles i.e. molar mass.

Slide3

Keys to Success

Make a plan and follow through.

Make sure

you have a complete balanced equation.

Make the correct mole-mass conversions.

Use the correct ratios.

Use the correct labels and units throughout.

Slide4

Steps to Solve Stoichiometric Problems.

1) Write and balance the equation.

2) Determine the

given

quantity and the

wanted

quantity.

3) If given is in grams, convert to moles, (divide by the molar mass).

4) Use the balanced equation to find the molar ratio between the

wanted

and the

given

.

5) Multiply the

given

by the molar ratio, (wanted/given ).

6) If the

wanted

is requested in moles, you are finished.

7) If the

wanted

is requested in grams, multiply by the molar mass.

Slide5

Steps to Solve Stoichiometric Problems.

Problem:

How many grams of silver sulfate would be produced from the reaction of 35.0 g of silver nitrate with an excess of magnesium sulfate?

Slide6

Steps to Solve Stoichiometric Problems.

How many grams of silver sulfate would be produced from the reaction of 35.0 g of silver nitrate with an excess of magnesium sulfate?

1) Write and balance the equation

.

2

AgNO

3

+ MgSO

4

 Ag

2

SO

4

+ Mg(NO

3

)

2

Slide7

Steps to Solve Stoichiometric Problems.

How many grams of silver sulfate would be produced from the reaction of 35.0 g of silver nitrate with an excess of magnesium sulfate?

2) Determine the

given

quantity and the

wanted

quantity.

Wanted: _______ g Ag

2

SO

4

Given: 35.0 g AgNO

3

Slide8

Steps to Solve Stoichiometric Problems.

How many grams of silver sulfate would be produced from the reaction of 35.0 g of silver nitrate with an excess of magnesium sulfate?

3) If given is in grams, convert to moles, (divide by the molar mass).

35.0 g AgNO

3

/169.9 g/

mol

= 0.206

mol

AgNO

3

Slide9

Steps to Solve Stoichiometric Problems.

How many grams of silver sulfate would be produced from the reaction of 35.0 g of silver nitrate with an excess of magnesium sulfate?

4) Use the balanced equation to find the molar ratio between the

wanted

and the

given

.

1

mol

Ag

2

SO

4

: 2

mol

AgNO

3

Slide10

Steps to Solve Stoichiometric Problems.

How many grams of silver sulfate would be produced from the reaction of 35.0 g of silver nitrate with an excess of magnesium sulfate?

5) Multiply the

given

by the molar ratio, (wanted/given ).

0.206

mol

AgNO

3

x ( 1mol Ag

2

SO

4

/ 2

mol

AgNO

3

) =

0.103

mol

Ag

2

SO

4

Slide11

Steps to Solve Stoichiometric Problems.

How many grams of silver sulfate would be produced from the reaction of 35.0 g of silver nitrate with an excess of magnesium sulfate?

6) If the

wanted

is requested in moles, you are finished.

Wanted in grams

 continue to Step 7

Slide12

Steps to Solve Stoichiometric Problems.

How many grams of silver sulfate would be produced from the reaction of 35.0 g of silver nitrate with an excess of magnesium sulfate?

7) If the

wanted

is requested in grams, multiply by the molar mass.

0.103

mol

Ag

2

SO

4

x 311.9 g/

mol

=

32.13 g Ag

2

SO

4