The mass of iron oxide formed by heating 53.5 g of iron (III) hydroxide is

n (iron (III) hydroxide) = m / M = 53.5 / 107 = 0.5 mol
the number of moles of iron (III) oxide = half the number of moles of iron (III) hydroxide, which means
m (oxide) = M (oxide) * n (hydroxide) / 2 = 160 * 0.25 = 40 g

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