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

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

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Answer 1 · 2015-11-15T21:16:21

$2.5 \times N_{A}$ $2.5 xx N_A$ , where $N_{A}$ $N_A$ is Avogadro's number, $6.02214 \times 10^{23}$ $6.02214 xx 10^23$ .

Explanation:

$N_{A}$ $N_A$ 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 $A l$ $Al$ atoms, I have a mass of $26.98$ $26.98$ $g$ $g$ of $A l$ $Al$ . You have two and a half times this number of $A l$ $Al$ atoms. What is the mass of your aluminum??

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

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