Question

How many grams of Al2O3 can be formed when 3.33 moles of Al and 4.87 moles of O2 react according to the following balanced equation: 4 Al + 3 O2 --> 2 Al2O3 (Hint: find the limiter first!) Express the answer to 3 sig figs. Do NOT include units.

Answer value

Answer 1

Answer: 170 grams of
`Al_2O_3`

Step-by-step explanation:

The balanced chemical reaction is:

`4Al+3O_2(g)\rightarrow 2Al_2O_3`

According to stoichiometry :

4 moles of
`Al` require = 3 moles of
`O_2`

Thus 3.33 moles of
`Al` will require=
`(3)/(4)* 3.33=2.49moles` of
`O_2`

Thus
`Al` is the limiting reagent as it limits the formation of product and
`O_2` is the excess reagent.

As 4 moles of
`Al` give = 2 moles of
`Al_2O_3`

Thus 3.33 moles of
`Al` give =
`(2)/(4)* 3.33=1.67moles` of
`Al_2O_3`

Mass of
`Al_2O_3=moles* {\text {Molar mass}}=1.67moles* 101.96g/mol=170g`

Thus 170 g of
`Al_2O_3` will be produced.