How many grams of sodium bicarbonate (NaHCO_3) are required to produce 40 mL of CO_2 gas at standard temperature and pressure (STP)?

• A. 0.42 g
• B. 0.21 g
• C. 0.084 g
• D. 0.0084 g

Answer: (C)

Verified step by step guidance

1. Identify the chemical reaction involved: Sodium bicarbonate (NaHCO\_3) decomposes to produce carbon dioxide (CO\_2) gas. The balanced chemical equation is: \(\mathrm{2\ NaHCO\_3 \rightarrow Na\_2CO\_3 + CO\_2 + H\_2O}\).
2. Determine the volume of CO\_2 gas given, which is 40 mL. Convert this volume to liters because gas volumes in stoichiometry are typically in liters: \$40\ \mathrm{mL} = 0.040\ \mathrm{L}$.
3. Use the molar volume of a gas at STP to find the moles of CO\_2 produced. At STP, 1 mole of any gas occupies 22.4 L. Calculate moles of CO\_2 using: \(\mathrm{moles\ of\ CO\_2 = \frac{volume\ of\ CO\_2}{22.4\ L/mol}}\).
4. From the balanced equation, 2 moles of NaHCO\_3 produce 1 mole of CO\_2. Use this mole ratio to find moles of NaHCO\_3 required: \(\mathrm{moles\ of\ NaHCO\_3 = 2 \times moles\ of\ CO\_2}\).
5. Calculate the mass of NaHCO\_3 needed by multiplying the moles of NaHCO\_3 by its molar mass (approximately 84 g/mol): \(\mathrm{mass\ of\ NaHCO\_3 = moles\ of\ NaHCO\_3 \times 84\ g/mol}\).