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?

Write a complete and balanced equation.

2Al2O3 + 6Br2 --------> 4AlBr3 + 3O2

Draw columns after each chemical, and divide your paper in half.

2Al2O3 + 6Br2 --------> 4AlBr3 + 3O2

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

25.42g

17.65g

Convert the given into moles.

2Al2O3 + 6Br2 --------> 4AlBr3 + 3O2

25.42g*1mole/159.808=

.1591moles

17.65g*1mole/101.961=

.1731moles

In each of the other columns, write moles given in each of the levels times a fraction.

2Al2O3 + 6Br2 --------> 4AlBr3 + 3O2

.1591* /25.42g*1mole/159.808=

.1591moles .1591* / .1591* /

17.65g*1mole/101.961=

.1731moles .1731* / .1731* / .1731* /

The numerator of the fraction is the coefficient of that column.

2Al2O3 + 6Br2 --------> 4AlBr3 + 3O2

.1591* 2/25.42g*1mole/159.808=

.1591moles .1591* 4/ .1591* 3/

17.65g*1mole/101.961=

.1731moles .1731* 6/ .1731* 4/ .1731* 3/

The denominator of the fraction is the coefficient of the given column on the same

level. (top or bottom)

2Al2O3 + 6Br2 --------> 4AlBr3 + 3O2

.1591* 2/625.42g*1mole/159.808=

.1591moles .1591* 4/6 .1591* 3/6

17.65g*1mole/101.961=

.1731moles .1731* 6/2 .1731* 4/2 .1731* 3/2

Do math to find moles.

2Al2O3 + 6Br2 --------> 4AlBr3 + 3O2

.1591* 2/6=

.0530moles

25.42g*1mole/159.808=

.1591moles

.1591* 4/6=

.1061moles

.1591* 3/6=

.0796moles

17.65g*1mole/101.961=

.1731moles.1731* 6/2=

.5193moles

.1731* 4/2=

.3462moles

.1731* 3/2=

.2597moles

Compare moles of one element. Cross off the row with more moles.

2Al2O3 + 6Br2 --------> 4AlBr3 + 3O2

.1591* 2/6=

.0530moles

25.42g*1mole/159.808=

.1591moles

.1591* 4/6=

.1061moles

.1591* 3/6=

.0796moles

17.65g*1mole/101.961=

.1731moles.1731* 6/2=

.5193moles

.1731* 4/2=

.3462moles

.1731* 3/2=

.2597moles

Convert moles into grams.

2Al2O3 + 6Br2 --------> 4AlBr3 + 3O2

.1591* 2/6=

.0530moles*101.961g/\

1mole=

5.404g

25.42g*1mole/159.808=

.1591moles*159.808g/

1mole=

25.43g

.1591* 4/6=

.1061moles*266.694g/

1mole=

28.30g

.1591* 3/6=

.0796moles*31.998g/

1mole=

2.547g

17.65g*1mole/101.961=

.1731moles.1731* 6/2=

.5193moles

.1731* 4/2=

.3462moles

.1731* 3/2=

.2597moles

Verify law of conservation of mass.

2Al2O3 + 6Br2 --------> 4AlBr3 + 3O2

.1591* 2/6=

.0530moles*101.961g/\

1mole=

5.404g

25.42g*1mole/159.808=

.1591moles*159.808g/

1mole=

25.43g

=30.83g

.1591* 4/6=

.1061moles*266.694g/

1mole=

28.30g

.1591* 3/6=

.0796moles*31.998g/

1mole=

2.547g

=30.85g

17.65g*1mole/101.961=

.1731moles.1731* 6/2=

.5193moles

.1731* 4/2=

.3462moles

.1731* 3/2=

.2597moles

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.