0 g of Mg, we need to use stoichiometry. From the balanced equation, we know that 2 moles of Mg react with 1 mole of $\mathrm{O}_{2}$. The molar mass of Mg is 24.31 g/mol, so 25.0 g of Mg is equal to:
$$
25.0 \mathrm{g} \times \frac{1 \mathrm{mol}}{24.31
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