1 / 14

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?.

miya
Télécharger la présentation

Stoichiometry with two givens

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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.

More Related