Question

If 1/2 moles of oxygen combine with aluminium to form Al2O3then weight of aluminium metal used in the reactions is

Answer 1

Answer: The mass of aluminium used in the reaction is 17.982 grams.

Explanation:

For the reaction of aluminium and oxygen to form aluminium oxide, the chemical reaction follows:

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

By stoichiometry of the reaction:

3 moles of oxygen gas reacts with 4 moles of aluminium.

So, $\frac{1}{2}$ mole of oxygen gas will react with $=\frac{4}{3}\times \frac{1}{2}=\frac{2}{3}=0.666mol$ of aluminium metal.

To calculate the mass of aluminium, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of aluminium = 27 g/mol

Putting values in above equation, we get:

$0.666mol=\frac{\text{Mass of aluminium}}{27g/mol}\\\\\text{Mass of aluminium}=17.982g$

Hence, the mass of aluminium used in the reaction is 17.982 grams.

Answer 2

Answer:

18g

Explanation:

balanced eq. =   4Al2  +  3O2 -------> 2Al2O3

3 mole of O2 combine with 4 mole of Al

1/2  mole will combine with

                      4     X     1    =   2      moles of Al              4   is S.C of Al

                       2            2           3                                       2  is S.C of O2

Mass of Al ==     moles of Al X atomic mass = mass of Al

                                         4   X 27 = 18 g

                                          2

Mass of Al = 18g