Step 1: Find moles of nitrate ion.
We have $100\,\text{mL}$ of $0.1\,M$ $NO_3^-$. \[ \text{moles of } NO_3^- = 0.1 \times \frac{100}{1000} = 0.01\,\text{mol} \]
Step 2: Find the change in oxidation state of nitrogen.
In $NO_3^-$ nitrogen is $+5$ since $x + 3(-2) = -1$ gives $x = +5$. In $NH_3$ nitrogen is $-3$ since $x + 3(+1) = 0$ gives $x = -3$. So each nitrogen gains \[ 5 - (-3) = 8 \text{ electrons} \]
Step 3: Find the change in oxidation state of magnesium.
Magnesium goes from $0$ in $Mg$ to $+2$ in $Mg^{2+}$, losing $2$ electrons per atom.
Step 4: Balance the electrons exchanged.
Total electrons gained by nitrate must equal total lost by magnesium. Electrons gained $= 0.01 \times 8 = 0.08\,\text{mol}$ of electrons. Moles of $Mg$ needed \[ = \frac{0.08}{2} = 0.04\,\text{mol} \]
Step 5: Convert moles of magnesium to grams.
The molar mass of magnesium is $24\,\text{g/mol}$. \[ \text{mass} = 0.04 \times 24 = 0.96\,\text{g} \]
Step 6: State the answer.
Therefore $0.96\,\text{g}$ of magnesium is required, matching the key.
\[ \boxed{0.96\,\text{g}} \]