Question

calculate the number of iron atoms in a 70.0 g sample of iron(iii) oxide (fe₂o₃).
be sure your answer has a unit symbol if necessary, and round it to 3 significant digits.

Explanation:

Step1: Calculate the molar mass of \(Fe_2O_3\)

The molar mass of \(Fe\) is \(55.85\ g/mol\) and of \(O\) is \(16.00\ g/mol\).
For \(Fe_2O_3\), \(M = 2\times55.85+3\times16.00=111.7 + 48.00=159.7\ g/mol\)

Step2: Calculate the number of moles of \(Fe_2O_3\)

Using the formula \(n=\frac{m}{M}\), where \(m = 70.0\ g\) and \(M = 159.7\ g/mol\)
\(n=\frac{70.0}{159.7}\approx0.438\ mol\)

Step3: Determine the number of moles of \(Fe\) atoms

In one molecule of \(Fe_2O_3\), there are \(2\) \(Fe\) atoms. So, moles of \(Fe\) \(n_{Fe}=2\times n_{Fe_2O_3}\)
\(n_{Fe}=2\times0.438 = 0.876\ mol\)

Step4: Calculate the number of \(Fe\) atoms

Using Avogadro's number \(N_A=6.022\times 10^{23}\ atoms/mol\)
\(N = n_{Fe}\times N_A\)
\(N=0.876\times6.022\times 10^{23}\)
\(N = 5.2852\times10^{23}\approx5.29\times 10^{23}\)

Answer:

\(5.29\times 10^{23}\)