luminum and oxygen react according to the following equation: 4Al(s) +3O2(g) --> 2Al2O3(s) What mass of Al2O3, in grams, can be made by reacting 4.6 g Al with excess oxygen?

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{4.6g}{27g/mol}=0.17moles[/tex]

[tex]4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)[/tex]

As oxygen is in excess, Aluminium is the limiting reagent and limits the formation of products.

According to stoichiometry:

4 moles of aluminium give = 2 moles of [tex]Al_2O_3(s)[/tex]

Thus 0.17 moles of aluminium give=[tex]\frac{2}{4}\times 0.17=0.085mol[/tex]

Mass of [tex]Al_2O_3=moles\times {\text {molar mass}}=0.085\times 102g/mol=8.7g[/tex]

Thus the mass of [tex]Al_2O_3(s)[/tex] is 8.7 grams