# Aluminum metal reacts with a solution of copper (II) nitrate. If you want to produce 5.4 * 10^24 atoms of copper, how many moles of aluminum metal are needed?

Jan 31, 2016

$5.98$ mols of aluminium metal are need to produce 5.4⋅10^24
atoms of copper.

#### Explanation:

First we should always have a balanced chemical equation for any reaction we are calculating.

The balanced chemical equation for the reaction of aluminum metal with a solution of copper (II) nitrate is as follows:

$2 A l + 3 C u {\left(N {o}_{3}\right)}_{2} \to 2 A l {\left(N {o}_{3}\right)}_{3} + 3 C u$

Secondly we should allways know the conversion between atoms and mols. 1 mol = $6.022 \cdot {10}^{23}$.

The copper has 5.4⋅10^24 atoms. We can convert this to mols by:

$\frac{5.4 \cdot {10}^{24}}{6.022 \cdot {10}^{23}}$

= $8.967120558$ mols
= $8.97$ mols

As the coefficients (the numbers at the front of the compounds in the chemical equation) are different to each other, we must change the mols.

As $3 C u$ = $8.97$ mols
So the aluminium must be $2 A l$ = $\left(\frac{8.97}{3}\right) \cdot 2$
$2 A l$ = $5.98$ mol

$\therefore$ $5.98$ mols of aluminium metal are need to produce 5.4⋅10^24
atoms of copper.