Iron (III) hydroxide weighing 53.5 g was treated with a solution of sulfuric acid. The mass of the resulting average salt is
n (hydroxide) = m / M = 53.5 / 107 = 0.5 mol
the number of moles of iron (III) sulfate = half the number of moles of iron (III) hydroxide, which means
m (salt) = M (iron (III) sulfate) * n (hydroxide) * 0.5 = 100 g
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