Elixirs such as Alka-Seltzer use the reaction of sodium bicarbonate with citric acid in aqueous solution to produce a fizz: $$ 3 \mathrm{NaHCO}_{3}(a q)+\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}(a q) \longrightarrow $$ $$ 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q) $$ a. What mass of \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}\) should be used for every \(1.0 \times 10^{2} \mathrm{mg} \mathrm{NaHCO}_{3} ?\) b. What mass of \(\mathrm{CO}_{2}(g)\) could be produced from such a mixture?

First, let's write down the balanced chemical equation of the reaction given in the exercise: \[3 \mathrm{NaHCO}_{3}(a q)+\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}(a q) \longrightarrow 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q)\]

Next, we need to find the molar masses of sodium bicarbonate (\(\mathrm{NaHCO}_{3}\)), citric acid (\(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}\)) and carbon dioxide (\(\mathrm{CO}_{2}\)). Using the periodic table, we can find the molar masses as follows: Molar mass of \(\mathrm{NaHCO}_{3} = 23 + 1 + 12 + (3 \times 16) = 84 \, \mathrm{g/mol}\) Molar mass of \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7} = (6 \times 12) + (8 \times 1) + (7 \times 16) = 192 \,\mathrm{g/mol}\) Molar mass of \(\mathrm{CO}_{2} = 12 + (2\times16) = 44 \, \mathrm{g/mol}\)

We know the mass of sodium bicarbonate to be used: \(1.0 \times 10^{2} \mathrm{mg} \mathrm{NaHCO}_{3} = 0.100 \,\mathrm{g}\) (converting mg to g). Now, we need to determine the stoichiometric ratio of citric acid to sodium bicarbonate from the balanced equation. The ratio is \(\frac{\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}}{\mathrm{NaHCO}_{3}} = \frac{1}{3}\). Using the given mass of \(\mathrm{NaHCO}_{3}\) and their molar masses, we can calculate the mass of citric acid required: Mass of \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7} = \) (Mass of \(\mathrm{NaHCO}_{3}) \times \frac{\text{Molar mass of}\, \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}}{\text{Molar mass of}\, \mathrm{NaHCO}_{3}} \times\frac{\text{Mole ratio of}\, \mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}}{\text{Mole ratio of}\, \mathrm{NaHCO}_{3}}\) Mass of \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7} = 0.100 \,\mathrm{g} \times \frac{192 \,\mathrm{g/mol}}{84 \, \mathrm{g/mol}} \times \frac{1}{3} = 0.0762 \, \mathrm{g}\)

Using a similar procedure, we'll find the mass of \(\mathrm{CO}_{2}\) produced from the same mass of sodium bicarbonate. The stoichiometric ratio of \(\mathrm{CO}_{2}\) to \(\mathrm{NaHCO}_{3}\) is \(\frac{3}{3} = 1.\) Mass of \(\mathrm{CO}_{2} = \) (Mass of \(\mathrm{NaHCO}_{3}) \times \frac{\text{Molar mass of}\, \mathrm{CO}_{2}}{\text{Molar mass of}\, \mathrm{NaHCO}_{3}} \times\frac{\text{Mole ratio of}\, \mathrm{CO}_{2}}{\text{Mole ratio of}\, \mathrm{NaHCO}_{3}}\) Mass of \(\mathrm{CO}_{2} = 0.100 \,\mathrm{g} \times \frac{44 \,\mathrm{g/mol}}{84 \, \mathrm{g/mol}} \times \frac{3}{3} = 0.0524 \, \mathrm{g}\) Now we have our results: a. For every \(1.0 \times 10^{2} \mathrm{mg} \mathrm{NaHCO}_{3}\), \(0.0762 \, \mathrm{g}\) of citric acid (\(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}\)) should be used. b. From such a mixture, \(0.0524 \, \mathrm{g}\) of carbon dioxide (\(\mathrm{CO}_{2}\)) could be produced.