# You need 2.5 moles of aluminum for an experiment. How many atoms of aluminum is this?

$2.5 \times {N}_{A}$, where ${N}_{A}$ is Avogadro's number, $6.02214 \times {10}^{23}$.
${N}_{A}$ is simply a number, admittedly a very large number. When I have such a number of any element/molecule I have an equivalent mass, and an equivalent number of atoms or molecules. When I have such a number of $A l$ atoms, I have a mass of $26.98$ $g$ of $A l$. You have two and a half times this number of $A l$ atoms. What is the mass of your aluminum??