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Stoichiometry

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.

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Stoichiometry

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  1. Stoichiometry By Luke Sayers

  2. Stoichiometry What is the mass of all the chemicals in a reaction between Potassium Hydroxide and Aluminum when 4.18g of Aluminum is used?

  3. 1.Write a complete and balanced chemical equation 3K(OH) + Al --> Al(OH)3 + 3K

  4. 2.Draw a column for each chemical 3K(OH) + Al --> Al(OH)3 + 3K

  5. 3. Write the amount of given in the appropriate column 3K(OH) + Al --> Al(OH)3 + 3K 4.18g

  6. 4.Convert the amount given into moles 3K(OH) + Al --> Al(OH)3 + 3K 4.18g 4.18g/1 x 1/26.981= .115mol

  7. 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

  8. 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

  9. 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

  10. 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

  11. 6.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

  12. 7.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

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