Question

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?

Answer 1

Answer: 8.7 grams

Explanation:

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

To calculate the moles, we use the equation:

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

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

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 $Al_2O_3(s)$

Thus 0.17 moles of aluminium give=$\frac{2}{4}\times 0.17=0.085mol$

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

Thus the mass of $Al_2O_3(s)$  is 8.7 grams