Answer: D. Each reagent contains one mole of a magnesium species per mole of magnesium sulfate formed (Mg, MgO and MgCO₃ each give one MgSO₄).
Reaction 1: \(n(\text{Mg}) = 1.2 / 24 = 0.050\,\text{mol}\). Reaction 2: \(n(\text{MgO}) = 2.0 / 40 = 0.050\,\text{mol}\). Reaction 3: \(n(\text{MgCO}_3) = 4.2 / 84 = 0.050\,\text{mol}\).
All three contain 0.050 mol, so all three reactions form the same number of moles of magnesium sulfate (the acid is in excess in every case). The masses were chosen so the moles come out equal.
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