Question

the rust that appears on steel surfaces is iron(iii) oxide. if the rust found spread over the surfaces of a steel bicycle frame contains a total of 2.806×10²³ oxygen atoms, how many grams of rust are present on the bicycle frame?

Explanation:

Step1: Determine the number of moles of oxygen atoms

We use Avogadro's number ($N_A = 6.022\times10^{23}\ atoms/mol$). The number of moles of oxygen atoms ($n_O$) is given by $n_O=\frac{N_O}{N_A}$, where $N_O = 2.806\times 10^{23}$ atoms. So, $n_O=\frac{2.806\times 10^{23}\ atoms}{6.022\times10^{23}\ atoms/mol}= 0.466\ mol$.

Step2: Find the number of moles of iron(III) oxide ($Fe_2O_3$)

In one - mole of $Fe_2O_3$, there are 3 moles of oxygen atoms. Let the number of moles of $Fe_2O_3$ be $n_{Fe_2O_3}$. Then $n_{Fe_2O_3}=\frac{n_O}{3}$. Substituting $n_O = 0.466\ mol$, we get $n_{Fe_2O_3}=\frac{0.466\ mol}{3}=0.1553\ mol$.

Step3: Calculate the mass of iron(III) oxide

The molar mass of $Fe_2O_3$ ($M_{Fe_2O_3}$) is calculated as follows: The molar mass of $Fe$ is $55.85\ g/mol$ and of $O$ is $16.00\ g/mol$. So, $M_{Fe_2O_3}=2\times55.85\ g/mol + 3\times16.00\ g/mol=111.7\ g/mol+48.00\ g/mol = 159.7\ g/mol$. The mass of $Fe_2O_3$ ($m_{Fe_2O_3}$) is given by $m_{Fe_2O_3}=n_{Fe_2O_3}\times M_{Fe_2O_3}$. Substituting $n_{Fe_2O_3}=0.1553\ mol$ and $M_{Fe_2O_3}=159.7\ g/mol$, we get $m_{Fe_2O_3}=0.1553\ mol\times159.7\ g/mol\approx23.4\ g$.

Answer:

$23.4\ g$