Final answer:
The mass of 6.5 x 10^24 molecules of Iron(III) Nitrate, Fe(NO3)3, is approximately 2609.63 grams.
Step-by-step explanation:
To find the mass of 6.5 x 10^24 molecules of Iron(III) Nitrate, Fe(NO3)3, using the factor-label method, we need to determine the molar mass of the compound and then multiply it by the number of molecules. The molar mass of Fe(NO3)3 is found by adding the atomic masses of each element:
Fe: 1 x 55.85 = 55.85 g/mol
N: 3 x 14.01 = 42.03 g/mol
O: 9 x 16.00 = 144.00 g/mol
Adding these values, we get 55.85 + 42.03 + 144.00 = 241.88 g/mol.
Now we can calculate the mass of 6.5 x 10^24 molecules of Fe(NO3)3. We use Avogadro's number (6.022 x 10^23 molecules/mol) to convert the number of molecules to moles:
6.5 x 10^24 molecules / (6.022 x 10^23 molecules/mol) = 10.794 mol
Finally, we multiply the number of moles by the molar mass:
10.794 mol x 241.88 g/mol = 2609.63 g.
Rounding to the nearest hundredth, the mass of 6.5 x 10^24 molecules of Fe(NO3)3 is approximately 2609.63 grams.