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Stoichiometry with two givens

Stoichiometry with two givens. By Erica Dougherty. If 17.65 g of aluminum oxide is reacted with 25.42 g of bromine, how many grams of aluminum bromide is produced? What is the excess reagent? What is the limiting reagent? How much of the excess reagent is left over?.

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Stoichiometry with two givens

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  1. Stoichiometry with two givens By Erica Dougherty

  2. If 17.65 g of aluminum oxide is reacted with 25.42 g of bromine, how many grams of aluminum bromide is produced? What is the excess reagent? What is the limiting reagent? How much of the excess reagent is left over?

  3. Write a complete and balanced equation. 2Al2O3 + 6Br2 --------> 4AlBr3 + 3O2

  4. Draw columns after each chemical, and divide your paper in half. 2Al2O3 + 6Br2 --------> 4AlBr3 + 3O2

  5. Write the amounts given in the proper columns, make sure one is on the top level and the other is on the bottom level. 2Al2O3 + 6Br2 --------> 4AlBr3 + 3O2 17.65g

  6. Convert the given into moles. 2Al2O3 + 6Br2 --------> 4AlBr3 + 3O2 17.65g*1mole/101.961= .1731moles

  7. In each of the other columns, write moles given in each of the levels times a fraction. 2Al2O3 + 6Br2 --------> 4AlBr3 + 3O2 17.65g*1mole/101.961= .1731moles .1731* / .1731* / .1731* /

  8. The numerator of the fraction is the coefficient of that column. 2Al2O3 + 6Br2 --------> 4AlBr3 + 3O2 17.65g*1mole/101.961= .1731moles .1731* 6/ .1731* 4/ .1731* 3/

  9. The denominator of the fraction is the coefficient of the given column on the same level. (top or bottom) 2Al2O3 + 6Br2 --------> 4AlBr3 + 3O2 17.65g*1mole/101.961= .1731moles .1731* 6/2 .1731* 4/2 .1731* 3/2

  10. Do math to find moles. 2Al2O3 + 6Br2 --------> 4AlBr3 + 3O2 17.65g*1mole/101.961= .1731moles .1731* 6/2= .5193moles .1731* 4/2= .3462moles .1731* 3/2= .2597moles

  11. Compare moles of one element. Cross off the row with more moles. 2Al2O3 + 6Br2 --------> 4AlBr3 + 3O2 17.65g*1mole/101.961= .1731moles .1731* 6/2= .5193moles .1731* 4/2= .3462moles .1731* 3/2= .2597moles

  12. Convert moles into grams. 2Al2O3 + 6Br2 --------> 4AlBr3 + 3O2 17.65g*1mole/101.961= .1731moles .1731* 6/2= .5193moles .1731* 4/2= .3462moles .1731* 3/2= .2597moles

  13. Verify law of conservation of mass. 2Al2O3 + 6Br2 --------> 4AlBr3 + 3O2 17.65g*1mole/101.961= .1731moles .1731* 6/2= .5193moles .1731* 4/2= .3462moles .1731* 3/2= .2597moles

  14. Answers • 28.30g of aluminum bromide are produced. • Aluminum oxide is the excess reagent. • Bromine is the limiting reagent. • There is 12.25g of aluminum oxide left.

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