Chapter 5
Q. 5.6
Graded
Sodium bicarbonate (baking soda) is widely used to absorb odors inside refrigerators. When acid is added to baking soda, the following reaction occurs:
NaHCO_{3}(s) + H^{+}(aq) → Na^{+}(aq) + CO_{2}(g) + H_{2}O
All experiments here are performed with 2.45 M HCl and 12.75 g of NaHCO_{3} at 732 mm Hg and 38°C.
ⓐ If an excess of HCl is used, what volume of CO_{2} is obtained?
ⓑ If NaHCO_{3} is in excess, what volume of HCl is required to produce 2.65 L of CO_{2}?
ⓒ What volume of CO_{2} is produced when all the NaHCO_{3} is made to react with 50.0 mL of HCl?
ⓐ
| ANALYSIS | |
| pressure (732 mm Hg); temperature (38°C); mass of NaHCO_{3} (12.75 g) | Information given: |
| molar mass of NaHCO_{3} stoichiometric ratio: 1 NaHCO_{3} /1 CO_{2} |
Information implied: |
| volume of CO_{2} produced | Asked for: |
STRATEGY
1. Follow the flow chart in Figure 5.5.
2. Convert to appropriate units of pressure and temperature.
ⓑ
| ANALYSIS | |
| pressure (732 mm Hg); temperature (38°C); volume of CO_{2} produced (2.65 L); molarity of HCl (2.45 M) |
Information given: |
| H^{+} is the reacting species. HCl is the parent compound. stoichiometric ratio: 1 H^{+}/1 CO_{2} |
Information implied: |
STRATEGY
Follow the flowchart in Figure 5.5.
V_{CO_{2}} \underrightarrow{PV = nRT} n_{CO_{2}} \overset{stoichiometric}{\underset{ratio}{\longrightarrow }} n_{H^{+}} \overset{atom}{\underset{ratio}{\longrightarrow }} n_{HCl} \underrightarrow{M} V_{HCl}ⓒ
| ANALYSIS | |
| molarity of HCl (2.45 M); volume of HCl (50.0 mL); pressure (732 mm Hg); temperature (38°C) |
Information given: |
| H^{+} is the reacting species. HCl is the parent compound. stoichiometric ratios: 1 H^{+}/1 CO_{2}; 1 NaHCO_{3} /1 CO_{2} from part (a): mol NaHCO_{3} |
Information implied: |
STRATEGY
1. The presence of enough given data to calculate the number of moles of each reactant tells you that part (c) is a limiting reactant problem.
2. Follow the flow chart in Figure 5.5 to determine the number of moles of CO_{2} obtained if HCl is limiting. You can obtain the moles of CO_{2} if NaHCO_{3} is limiting from part (a).
3. Compare the moles of CO_{2} obtained using H^{+} as the limiting reactant to the moles of CO_{2} obtained using NaHCO_{3} as the limiting reactant. Choose the smaller number of moles of CO_{2}.
4. Use the ideal gas law to convert mol CO_{2} to the volume of CO_{2}.
Step-by-Step
Verified Solution
ⓐ
| mol CO_{2} (n) | 12.75 g NaHCO_{2} × \frac{1 mol}{84.01 g} × \frac{1 mol CO_{2}}{1 mol NaHCO_{3}} = 0.1518 |
| volume CO_{2} (V) | V = \frac{0.1518 mol × 0.0821 L · atm/mol · K × (273 + 38)K}{(732/760)atm} = 4.02 L |
ⓑ
| mol CO_{2} | n = \frac{2.65 L × (732/760)atm}{0.0821 L · atm/mol · K × (273 + 38)K} = 0.100 |
| mol HCl | 0.100 mol CO_{2} × \frac{1 mol H^{+}}{1 mol CO_{2}} × \frac{1 mol HCL}{1 mol H^{+}} = 0.100 |
| Volume HCl | \frac{0.100 mol HCl}{2.45 mol/L} = 0.0408 L = 40.8 mL |
ⓒ
| mol CO_{2}: NaHCO_{2} limiting | from part (a): 0.1518 mol NaHCO_{3} ×\frac{1 mol CO_{2}}{1 mol NaHCO_{3}} = 0.1518 |
| mol CO_{2}: HCl limiting | (0.0500 L × 2.45 mol/L) mol HCl ×\frac{1 mol H^{+}}{1 mol HCl} × \frac{1 mol CO_{2}}{1 mol H^{+}} = 0.122 mol |
| Theoretical yield of CO_{2} | 0.122 < 0.1518; 0.122 mol CO_{2} obtained |
| Volume CO_{2} | V = \frac{0.122 mol × 0.0821 L · atm/mol · K × (273 + 38)K}{(732/760)atm} = 3.25 L |
END POINT
1. When a problem comes in several parts, you may not need to use all the given information for each part.
2. You should also check to see whether you can use information that you obtained from the preceding parts for subsequent questions.