In an experiment, you added hydrochloric acid (HCl) to fully neutralize a solution of sodium carbonate (Na2CO3). If you started with 0.050 mol of Na2CO3, how many moles of HCl were required for complete neutralization?

• A. 0.050 mol
• B. 0.075 mol
• C. 0.10 mol
• D. 0.025 mol

Answer: (C)

Verified step by step guidance

1. Write the balanced chemical equation for the reaction between sodium carbonate (Na\_2CO\_3) and hydrochloric acid (HCl). The reaction is: \(\mathrm{Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2}\)
2. Identify the mole ratio between sodium carbonate and hydrochloric acid from the balanced equation. For every 1 mole of Na\_2CO\_3, 2 moles of HCl are required.
3. Use the given amount of sodium carbonate (0.050 mol) and multiply by the mole ratio to find the moles of HCl needed: \(\text{moles of HCl} = 0.050 \text{ mol Na}_2\text{CO}_3 \times \frac{2 \text{ mol HCl}}{1 \text{ mol Na}_2\text{CO}_3}\)
4. Calculate the product from the previous step to find the total moles of HCl required for complete neutralization (do not compute the final number here, just set up the expression).
5. Interpret the result as the amount of HCl needed to fully neutralize the given amount of sodium carbonate in the experiment.