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Stoichiometry. By Luke Sayers. Stoichiometry. What is the mass of all the chemicals in a reaction between Potassium Hydroxide and Aluminum when 4.18g of Aluminum is used?. 1.Write a complete and balanced chemical equation. 3K(OH) + Al --> Al(OH) 3 + 3K. - PowerPoint PPT Presentation
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StoichiometryStoichiometry
By Luke SayersBy Luke Sayers
StoichiometryStoichiometry
What is the mass of all the chemicals in a reaction between Potassium Hydroxide and Aluminum when 4.18g of Aluminum is used?
What is the mass of all the chemicals in a reaction between Potassium Hydroxide and Aluminum when 4.18g of Aluminum is used?
1.Write a complete and balanced chemical equation
1.Write a complete and balanced chemical equation
3K(OH) + Al --> Al(OH)3 + 3K3K(OH) + Al --> Al(OH)3 + 3K
2.Draw a column for each chemical2.Draw a column for each chemical
3K(OH) + Al --> Al(OH)3 + 3K3K(OH) + Al --> Al(OH)3 + 3K
3. Write the amount of given in the appropriate column
3. Write the amount of given in the appropriate column
3K(OH) + Al --> Al(OH)3 + 3K4.18g
3K(OH) + Al --> Al(OH)3 + 3K4.18g
4.Convert the amount given into moles4.Convert the amount given into moles
3K(OH) + Al --> Al(OH)3 + 3K4.18g
4.18g/1 x 1/26.981=
.115mol
3K(OH) + Al --> Al(OH)3 + 3K4.18g
4.18g/1 x 1/26.981=
.115mol
5.Find moles for each other chemicala) In each of the other columns write the moles of given (x) a
fraction
5.Find moles for each other chemicala) In each of the other columns write the moles of given (x) a
fraction
3K(OH) + Al --> Al(OH)3 + 3K.115mol x 4.18g .115mol x .115mol x
4.18g/1 x 1/26.981=
.115mol
3K(OH) + Al --> Al(OH)3 + 3K.115mol x 4.18g .115mol x .115mol x
4.18g/1 x 1/26.981=
.115mol
5.Find moles for each other chemicalb) The numerator of the fraction is the coefficient of that
column
5.Find moles for each other chemicalb) The numerator of the fraction is the coefficient of that
column
3K(OH) + Al --> Al(OH)3 + 3K.115mol x 3/ 4.18g .115mol x 1/ .115mol x 3/
4.18g/1 x 1/26.981=
.115mol
3K(OH) + Al --> Al(OH)3 + 3K.115mol x 3/ 4.18g .115mol x 1/ .115mol x 3/
4.18g/1 x 1/26.981=
.115mol
5.Find moles for each other chemicalc) The denominator of the fraction is the coefficient of the
given column
5.Find moles for each other chemicalc) The denominator of the fraction is the coefficient of the
given column
3K(OH) + Al --> Al(OH)3 + 3K.115mol x 3/1= 4.18g .115mol x 1/1= .115mol x
3/1=
4.18g/1 x 1/26.981=
.115mol
3K(OH) + Al --> Al(OH)3 + 3K.115mol x 3/1= 4.18g .115mol x 1/1= .115mol x
3/1=
4.18g/1 x 1/26.981=
.115mol
5.Find moles for each other chemicald) Do the math and label as moles
5.Find moles for each other chemicald) Do the math and label as moles
3K(OH) + Al --> Al(OH)3 + 3K.115mol x 3/1= 4.18g .115mol x 1/1= .115mol x
3/1=
.465mol 4.18g/1 x 1/26.981= .115mol .465mol
.115mol
3K(OH) + Al --> Al(OH)3 + 3K.115mol x 3/1= 4.18g .115mol x 1/1= .115mol x
3/1=
.465mol 4.18g/1 x 1/26.981= .115mol .465mol
.115mol
6.Convert all moles into grams6.Convert all moles into grams
3K(OH) + Al --> Al(OH)3 + 3K.115mol x 3/1= 4.18g .115mol x 1/1= .115mol x
3/1=
.465mol x 4.18g/1 x 1/26.981= .115mol x .465mol x
56.1056g/1 = .115mol 78.0034g/1 = 39.0983g/1 =
26.1g 12.1g 18.2g
3K(OH) + Al --> Al(OH)3 + 3K.115mol x 3/1= 4.18g .115mol x 1/1= .115mol x
3/1=
.465mol x 4.18g/1 x 1/26.981= .115mol x .465mol x
56.1056g/1 = .115mol 78.0034g/1 = 39.0983g/1 =
26.1g 12.1g 18.2g
7.Verify the Law of Conservation of mass7.Verify the Law of Conservation of mass
3K(OH) + Al --> Al(OH)3 + 3K.115mol x 3/1= 4.18g .115mol x 1/1= .115mol x
3/1=
.465mol x 4.18g/1 x 1/26.981= .115mol x .465mol x
56.1056g/1 = .115mol 78.0034g/1 = 39.0983g/1 =
26.1g 12.1g 18.2g
26.1g+4.18g=30.3g 12.1g+18.2g=30.3g
3K(OH) + Al --> Al(OH)3 + 3K.115mol x 3/1= 4.18g .115mol x 1/1= .115mol x
3/1=
.465mol x 4.18g/1 x 1/26.981= .115mol x .465mol x
56.1056g/1 = .115mol 78.0034g/1 = 39.0983g/1 =
26.1g 12.1g 18.2g
26.1g+4.18g=30.3g 12.1g+18.2g=30.3g
