Question 3.12: Alka-Seltzer tablets contain aspirin, sodium bicarbonate, an...
Alka-Seltzer tablets contain aspirin, sodium bicarbonate, and citric acid. When they come into contact with water, the sodium bicarbonate (NaHCO3) and citric acid (H3C6H5O7) react to form carbon dioxide gas, among other products.
3NaHCO3(aq) + H3C6H5O7(aq) → 3CO2(g) + 3H2O(l) + Na3C6H5O7(aq)
The formation of CO2 causes the trademark fizzing when the tablets are dropped into a glass of water. An Alka-Seltzer tablet contains 1.700 g of sodium bicarbonate and 1.000 g of citric acid. Determine, for a single tablet dissolved in water, (a) which ingredient is the limiting reactant, (b) what mass of the excess reactant is left over when the reaction is complete, and (c) what mass of CO2 forms.
Strategy Convert each of the reactant masses to moles. Use the balanced equation to write the necessary stoichiometric conversion factor and determine which reactant is limiting. Again, using the balanced equation, write the stoichiometric conversion factors to determine the number of moles of excess reactant remaining and the number of moles of CO2 produced. Finally, use the appropriate molar masses to convert moles of excess reactant and moles of CO2 to grams.
Setup The required molar masses are 84.01 g/mol for NaHCO3, 192.12 g/mol for H3C6H5O7, and 44.01 g/mol for CO2. From the balanced equation we have 3 mol NaHCO3 \bumpeq 1 mol H3C6H5O7, 3 mol NaHCO3 \bumpeq 3 mol CO2, and 1 mol H3C6H5O7 \bumpeq 3 mol CO2. The necessary stoichiometric conversion factors are therefore:
\frac{3 mol NaHCO_{3}}{1 mol H_{3}C_{6}H_{5}O_{7}} \frac{1 mol H_{3}C_{6}H_{5}O_{7}}{3 mol NaHCO_{3}} \frac{3 mol CO_{2}}{3 mol NaHCO_{3}} \frac{3 mol CO_{2}}{1 mol H_{3}C_{6}H_{5}O_{7}}
1.700 \cancel{g NaHCO_{3}} × \frac{1 mol NaHCO_{3}}{84.01 \cancel{g NaHCO_{3}}} = 0.02024 mol NaHCO_{3}
1.000 \cancel{g H_{3}C_{6}H_{5}O_{7}} × \frac{1 mol H_{3}C_{6}H_{5}O_{7}}{192.12 \cancel{g H_{3}C_{6}H_{5}O_{7}}} = 0.005205 mol H_{3}C_{6}H_{5}O_{7}
(a) To determine which reactant is limiting, calculate the amount of citric acid necessary to react completely with 0.02024 mol sodium bicarbonate.
0.02024 \cancel{mol NaHCO_{3}} × \frac{1 mol H_{3}C_{6}H_{5}O_{7}}{3 \cancel{mol NaHCO_{3}}} = 0.006745 mol H_{3}C_{6}H_{5}O_{7}
The amount of H3C6H5O7 required to react with 0.02024 mol of NaHCO3 is more than a tablet contains. Therefore, citric acid is the limiting reactant and sodium bicarbonate is the excess reactant.
(b) To determine the mass of excess reactant (NaHCO3) left over, first calculate the amount of NaHCO3 that will react:
0.005205 \cancel{mol H_{3}C_{6}H_{5}O_{7}} × \frac{3 mol NaHCO_{3}}{1 \cancel{mol H_{3}C_{6}H_{5}O_{7}}} = 0.01562 mol NaHCO_{3}
Thus, 0.01562 mole of NaHCO3 will be consumed, leaving 0.00462 mole unreacted. Convert the unreacted amount to grams as follows:
0.00462 \cancel{mol NaHCO_{3}} × \frac{84.01 g NaHCO_{3}}{1 \cancel{mol NaHCO_{3}}} = 0.388 g NaHCO_{3}
(c) To determine the mass of CO2 produced, first calculate the number of moles of CO2 produced from the number of moles of limiting reactant (H3C6H5O7) consumed:
0.005205 \cancel{mol H_{3}C_{6}H_{5}O_{7}} × \frac{3 mol CO_{2}}{1 \cancel{mol H_{3}C_{6}H_{5}O_{7}}} = 0.01562 mol CO_{2}
Convert this amount to grams as follows:
0.01562 \cancel{mol CO_{2}} × \frac{44.01 g CO_{2}}{1 \cancel{mol CO_{2}}} = 0.6874 g CO_{2}
To summarize the results: (a) citric acid is the limiting reactant, (b) 0.388 g sodium bicarbonate remains unreacted, and (c) 0.6874 g carbon dioxide is produced.