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## Q. 5.6

Sodium bicarbonate (baking soda) is widely used to absorb odors inside refrigerators. When acid is added to baking soda, the following reaction occurs:

NaHC$O_{3}$(s) + $H^{+}$(aq) → $Na^{+}$(aq) + C$O_{2}$(g) + $H_{2}O$

All experiments here are performed with 2.45 M HCl and 12.75 g of NaHC$O_{3}$ at 732 mm Hg and 38°C.

ⓐ If an excess of HCl is used, what volume of C$O_{2}$ is obtained?
ⓑ If NaHC$O_{3}$ is in excess, what volume of HCl is required to produce 2.65 L of C$O_{2}$?
ⓒ What volume of C$O_{2}$ is produced when all the NaHC$O_{3}$ is made to react with 50.0 mL of HCl?

 ANALYSIS pressure (732 mm Hg); temperature (38°C); mass of NaHC$O_{3}$ (12.75 g) Information given: molar mass of NaHC$O_{3}$ stoichiometric ratio: 1 NaHC$O_{3}$ /1 C$O_{2}$ Information implied: volume of C$O_{2}$ produced Asked for:

STRATEGY

1. Follow the flow chart in Figure 5.5.
2. Convert to appropriate units of pressure and temperature.

$mass_{NaHCO_{3}} \underrightarrow{MM} n_{NaHCO_{3}} \overset{stoichiometric}{\underset{ratio}{\longrightarrow }} n_{CO_{2}} \underrightarrow{PV = nRT}V_{CO_{2}}$

 ANALYSIS pressure (732 mm Hg); temperature (38°C); volume of C$O_{2}$ 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 C$O_{2}$ 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 C$O_{2}$; 1 NaHC$O_{3}$ /1 C$O_{2}$ from part (a): mol NaHC$O_{3}$ 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 C$O_{2}$ obtained if HCl is limiting. You can obtain the moles of C$O_{2}$ if NaHC$O_{3}$ is limiting from part (a).
3. Compare the moles of C$O_{2}$ obtained using $H^{+}$ as the limiting reactant to the moles of C$O_{2}$ obtained using NaHC$O_{3}$ as the limiting reactant. Choose the smaller number of moles of C$O_{2}$.
4. Use the ideal gas law to convert mol C$O_{2}$ to the volume of C$O_{2}$. ## Verified Solution

 mol C$O_{2}$ (n) 12.75 g NaHC$O_{2}$ × $\frac{1 mol}{84.01 g}$ × $\frac{1 mol CO_{2}}{1 mol NaHCO_{3}}$ = 0.1518 volume C$O_{2}$ (V) V = $\frac{0.1518 mol × 0.0821 L · atm/mol · K × (273 + 38)K}{(732/760)atm}$ = 4.02 L

 mol C$O_{2}$ n = $\frac{2.65 L × (732/760)atm}{0.0821 L · atm/mol · K × (273 + 38)K}$ = 0.100 mol HCl 0.100 mol C$O_{2}$ × $\frac{1 mol H^{+}}{1 mol CO_{2}}$ × $\frac{1 mol HCL}{1 mol H^{+}}$ = 0.100 Volume HCl $\frac{0.100 mol HCl}{2.45 mol/L}$ = 0.0408 L = 40.8 mL

 mol C$O_{2}$: NaHC$O_{2}$ limiting from part (a): 0.1518 mol NaHC$O_{3}$ ×$\frac{1 mol CO_{2}}{1 mol NaHCO_{3}}$ = 0.1518 mol C$O_{2}$: 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 C$O_{2}$ 0.122 < 0.1518; 0.122 mol C$O_{2}$ obtained Volume C$O_{2}$ 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.