What mass of $\ce{NaHCO3}$ $(\text{M=84})$ is required to completely neutralize $25.0$ mL of $0.125$ M $\ce{H2SO4}$?

$ \textbf{(A) }\text{0.131 g}\qquad\textbf{(B) }\text{0.262 g}\qquad\textbf{(C) }\text{0.525 g}\qquad\textbf{(D) }\text{1.05 g}\qquad$
To solve this problem, we need to determine the mass of sodium bicarbonate ($\ce{NaHCO3}$) required to neutralize a given volume and concentration of sulfuric acid ($\ce{H2SO4}$). We will follow these steps:

1. **Write the balanced chemical equation for the reaction:**
   \[
   \ce{H2SO4 + 2 NaHCO3 -> Na2SO4 + 2 CO2 + 2 H2O}
   \]
   This equation shows that one mole of sulfuric acid reacts with two moles of sodium bicarbonate.

2. **Calculate the number of moles of $\ce{H2SO4}$:**
   \[
   \text{Moles of } \ce{H2SO4} = \text{Volume (L)} \times \text{Molarity (M)}
   \]
   Given:
   \[
   \text{Volume} = 25.0 \, \text{mL} = 0.025 \, \text{L}
   \]
   \[
   \text{Molarity} = 0.125 \, \text{M}
   \]
   \[
   \text{Moles of } \ce{H2SO4} = 0.025 \, \text{L} \times 0.125 \, \text{M} = 0.003125 \, \text{moles}
   \]

3. **Determine the moles of $\ce{NaHCO3}$ required:**
   According to the balanced equation, 1 mole of $\ce{H2SO4}$ reacts with 2 moles of $\ce{NaHCO3}$.
   \[
   \text{Moles of } \ce{NaHCO3} = 2 \times \text{Moles of } \ce{H2SO4}
   \]
   \[
   \text{Moles of } \ce{NaHCO3} = 2 \times 0.003125 \, \text{moles} = 0.00625 \, \text{moles}
   \]

4. **Calculate the mass of $\ce{NaHCO3}$ required:**
   \[
   \text{Mass} = \text{Moles} \times \text{Molar Mass}
   \]
   Given the molar mass of $\ce{NaHCO3}$ is 84 g/mol:
   \[
   \text{Mass of } \ce{NaHCO3} = 0.00625 \, \text{moles} \times 84 \, \text{g/mol} = 0.525 \, \text{g}
   \]

The final answer is $\boxed{0.525 \, \text{g}}$