How many moles of ferric alum (NH₄)₂SO₄Fe₂(SO₄)₃.24H₂O can be made from the sample of Fe containing 0.0056 g of
it?

Moles of Fe = $\frac{0.0056}{56}={10}^{-4}\mathrm{mol}$

1 mol of alum = 2 mol of Fe

2 mol of Fe = 1 mol of alum

${10}^{-4}\mathrm{mol} \mathrm{of} \mathrm{Fe} = \frac{1}{2}\times {10}^{-4}\mathrm{mol}=0.5\times {10}^{-4}\mathrm{mol}$