To determine how many grams of oxygen will react completely with 27 grams of aluminum (Al), we first need to look at the balanced chemical equation for the reaction between aluminum and oxygen to form aluminum oxide ($Al_2O_3$). The balanced equation is:

$4Al + 3O_2 \rightarrow 2Al_2O_3$

This equation tells us that 4 moles of aluminum react with 3 moles of oxygen gas to produce 2 moles of aluminum oxide.

First, we need to determine the molar mass of aluminum (Al) and oxygen (O₂) to proceed with the calculations.

The molar mass of aluminum (Al) is approximately $27 \, g/mol$.

The molar mass of oxygen gas (O₂) is $32 \, g/mol$ (since the molar mass of an oxygen atom is approximately 16 g/mol, and $O_2$ has two oxygen atoms).

Given that we have 27 g of aluminum:

The number of moles of aluminum ($n_{Al}$) can be calculated as follows:

$n_{Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} = \frac{27 \, g}{27 \, g/mol} = 1 \, mol$

According to the balanced equation, 4 moles of Al react with 3 moles of O₂. Therefore, to find out how many moles of O₂ react with 1 mole of Al, we set a proportion:

$\frac{4 \, mol \, Al}{3 \, mol \, O_2} = \frac{1 \, mol \, Al}{x \, mol \, O_2}$

Solving for $x$, we find:

$x = \frac{3 \, mol \, O_2}{4 \, mol \, Al} = \frac{3}{4} \, mol \, O_2$

To find out the mass of oxygen that will react, we use the molar mass of oxygen:
$mass \, O_2 = n_{O_2} \times \text{molar mass of }\, O_2 = \frac{3}{4} \, mol \times 32 \, g/mol = 24 \, g$

Therefore, 27 g of aluminum will react completely with 24 g of oxygen. The correct answer is Option D.