# Question a279a

Jan 14, 2018

$\text{1100 g}$

#### Explanation:

The key here is the fact that $1$ mole of any element contains $6.022 \cdot {10}^{23}$ atoms of that element, as given by Avogaro's constant.

So if your sample contains $1$ mole of iron, it contains $6.022 \cdot {10}^{23}$ atoms of iron. Similarly, if a sample contains $1$ mole of potassium, it contains $6.022 \cdot {10}^{23}$ atoms of potassium.

Now, you know that iron has a molar mass of ${\text{55.845 g mol}}^{- 1}$. This means that every time your sample contains $1$ mole of iron, it will have a mass of $\text{55.845 g}$. This means that $6.022 \cdot {10}^{23}$ atoms of irion, the equivalent of $1$ mole of iron, havea mass of $\text{55.845 g}$.

Potassium has a molar mass of ${\text{39.0983 g mol}}^{- 1}$, which means that $1$ mole of potassium has a mass $\text{39.0983 g}$. This means that $6.022 \cdot {10}^{23}$ atoms of potassium have a mass of $\text{39.0983 g}$.

So, use this to find the number of atoms of potassium present in $\text{780 g}$ of potassium.

780 color(red)(cancel(color(black)("g"))) * (6.022 * 10^(23) quad "atoms K")/(39.0983 color(red)(cancel(color(black)("g")))) = 120.137 * 10^(23) quad "atoms K"#

Now all you have to do is to figure out how many grams of iron will contain $120.137 \cdot {10}^{23}$ atoms of iron.

$120.137 \cdot \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{{10}^{23}}}} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{atoms Fe"))) * "55.845 g"/(6.022 * color(blue)(cancel(color(black)(10^(23)))) color(red)(cancel(color(black)("atoms Fe")))) = color(darkgreen)(ul(color(black)("1100 g}}}}$

The answer is rounded to two sig figs, the number of sig figs you have for the mass of iron.

Notice that because $1$ mole of atoms of iron is heavier than $1$ mole of atoms of potassium, which is essentially saying that $1$ atom of iron is heavier than $1$ atom of potassium, the mass of iron that contains the same number of atoms as the number of atoms of potassium present in $\text{780 g}$ of potassium is $>$ $\text{780 g}$.