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Chapter 7: Problem 30

What is the mass in grams of a sample of \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) that contains \(3.59 \times 10^{23}\) sulfate ions, \(\mathrm{SO}_{4}^{2-}\) ? The molar mass of \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) is 399.91 \(\mathrm{g} / \mathrm{mol}\)

Short Answer

Expert verified
Therefore, the mass of the sample of Fe2(SO4)3 that contains \(3.59 \times 10^{23}\) sulfate ions is 79.6 grams.

Step by step solution

01

Calculate the number of units of Fe2(SO4)3

Given that the sample contains \(3.59 \times 10^{23}\) sulfate ions and each unit of Fe2(SO4)3 contains 3 sulfate ions, the number of units of Fe2(SO4)3 in the sample can be calculated by dividing the number of sulfate ions by 3. This gives \(3.59 \times 10^{23} / 3 = 1.20 \times 10^{23}\) units of Fe2(SO4)3.
02

Convert the number of units to moles

The number of moles is the number of units divided by Avogadro's number (\(6.022 \times 10^{23}\)). This gives \(1.20 \times 10^{23} / 6.022 \times 10^{23} = 0.199\) moles of Fe2(SO4)3.
03

Calculate the mass of the sample

The mass of the sample can be found by multiplying the number of moles by the molar mass. The molar mass of Fe2(SO4)3 is given as 399.91 g/mol, so the mass of the sample is \(0.199\) moles * 399.91 g/mol = 79.6 g.

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