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 9.62×1022 oxygen atoms, how many grams of rust are present on the bicycle frame?

Answer 1

The grams of rust present in the bicycle frame is
`x = 8.50 g`

Step-by-step explanation:

From the question we are told that

The number of oxygen atom contained is
`n = 9.62 *10^(22) \ atoms`

The molar mass of the compound is
`M_(Fe_3O_3) = 159.69 g`

At standard temperature and pressure the number of oxygen atom in one mole of iron(III) oxide is mathematically evaluated as

`N_o = 3 * Ne`

Where Ne is the avogadro's constant with a value
`N_e = 6.023 *10^(23) \ atoms`

So

`N_o= 3 * 6.023*10^(23) \ atoms`

`N_o= 1.8069*10^(24) \ atoms`

So

`1.8069*10^(24) \ atoms` is contained in
`M_(Fe_3O_3) = 159.69 g`

`9.62 *10^(22) \ atoms` is contained in x

So

`x = (159.69 * 9.62 *10^(22))/(1.8069 *10^(24))`

`x = 8.50 g`