Answer:
The grams of rust present in the bicycle frame is [tex]x = 8.50 g[/tex]
Explanation:
From the question we are told that
The number of oxygen atom contained is [tex]n = 9.62 *10^{22} \ atoms[/tex]
The molar mass of the compound is [tex]M_{Fe_3O_3} = 159.69 g[/tex]
At standard temperature and pressure the number of oxygen atom in one mole of iron(III) oxide is mathematically evaluated as
[tex]N_o = 3 * Ne[/tex]
Where Ne is the avogadro's constant with a value [tex]N_e = 6.023 *10^{23} \ atoms[/tex]
So
[tex]N_o= 3 * 6.023*10^{23} \ atoms[/tex]
[tex]N_o= 1.8069*10^{24} \ atoms[/tex]
So
[tex]1.8069*10^{24} \ atoms[/tex] is contained in [tex]M_{Fe_3O_3} = 159.69 g[/tex]
[tex]9.62 *10^{22} \ atoms[/tex] is contained in x
So
[tex]x = \frac{159.69 * 9.62 *10^{22}}{1.8069 *10^{24}}[/tex]
[tex]x = 8.50 g[/tex]
The grams of rust present on the bicycle frame is 8.501 grams.
As mentioned, the rust is spread over the surfaces of the steel bicycle frame and comprise a total of 9.62 × 10²² atoms of oxygen. The rust, which appears on the steel surfaces is iron(III) oxide or Fe₂O₃.
Now the molecular weight of Fe₂O₃ is 160 grams per mole. Therefore, in the 160 grams of iron oxide, 3 × 6.023 × 10²³ number of oxygen is present. Therefore, 9.62 × 10²² oxygen atoms will be present in,
[tex]Rust(grams) = \frac{160 * 9.62 * 10^{22} }{3 * 6.023 * 10^{23} } \\Rust (grams) = 8.501 grams[/tex]
Thus, 8.501 grams of rust is present on the bicycle frame.
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