Question

How many grams of oxygen are needed to produce 408g of Al2O3?

Answer 1

Answer:

Explanation

GIVEN

• Mass of Al2O3 =, 408g

• Molecular mass Al2O3 =,101,96 g/mol

• Molecular Mass Oxygen =,15,999 g/mol

We will consider the following balanced chemical equation that takes place :

`4Al(s)\text{ + 3O}_2(g)\text{ }\Rightarrow\text{ 2Al}_2O_3`

( i) Calculate the moles of Al2O3

`\begin{gathered} Moles\text{ = }\frac{Mass}{Molecular\text{ Mass }} \\ \text{ = }\frac{408g\text{ }}{101.96\text{ g}}*mol \\ \text{ =4.0moles of Al}_2O_3 \end{gathered}`

(ii) Determine moles of Oxygen from the stoichiometry in the balanced reaaction

3 moles Oxygen reacts and produce 2 moles Al2O3

So, x moles Oxygen willreact and produce 4 moles Al2O3

Therefore,

X moles O2 = (4Moles Al2O3 *3 moles Oxygen ) / 2 moles Oxygen

= 6 moles of O2

(iii) Determine Mass of Oxygen :

`\begin{gathered} Mass\text{ O}_2=Moles\text{ O}_2*Molecular\text{ mass O}_2 \\ \text{ = 6 moles * 15.999g/moles} \\ \text{ =95.99 grams of O}_2 \end{gathered}`

Therefore, 95.99 grams of O2 are needed to produce 408g of Al2O3