Answer:
399.88 [tex]\frac{4.05*10^{23} }{6.022*10^{23} }[/tex] =269
Explanation:
399.88 is the molar mass of iron(III) sulfate.
The mass of 4.05 × [tex]10^{23[/tex] formula units of Fe2(SO4)3 is approximately 6.48 × [tex]10^{25[/tex] grams.
Mass from formula unitsTo calculate the molar mass of Fe2(SO4)3, we need to determine the atomic masses of the individual elements involved and multiply them by their respective subscripts:
Fe: 55.845 g/mol (atomic mass of iron)S: 32.06 g/mol (atomic mass of sulfur)O: 16.00 g/mol (atomic mass of oxygen)Fe2(SO4)3 consists of 2 iron atoms, 3 sulfur atoms, and 12 oxygen atoms:
Molar mass of Fe2(SO4)3 = (2 * 55.845) + (3 * 32.06) + (12 * 16.00) = 159.69 g/mol
To calculate the mass of 4.05 × 10^23 formula units of Fe2(SO4)3, we can use the molar mass:
Mass = (4.05 × 10^23) * (159.69 g/mol) = 6.48 × 10^25 g
Rounded to two decimal places, the mass of 4.05 × [tex]10^{23[/tex] formula units of Fe2(SO4)3 is approximately 6.48 × [tex]10^{25[/tex] grams.
More on formula units can be found here: https://brainly.com/question/33740227
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