Question

The fizz produced when an Alka-Seltzer tablet is dissolved in water is due to the reaction between sodium bicarbonate (NaHCO₃) and citric acid (H₃C₆H₅O₇):
3NaHCO₃(aq) + H₃C₆H₅O₇(aq) --------> 3CO₂(g) + 3H₂O(l) + Na₃C₆H₅O₇(aq)
In a certain experiment 1.00g of sodium bicarbonate and 1.00g of citric acid are allowed to react.
How many grams of carbon dioxide form?

Answer 1

Answer:

0.47g of $CO_{2}$

Explanation:

1. Balanced equation:

3NaHCO₃(aq) + H₃C₆H₅O₇(aq) --------> 3CO₂(g) + 3H₂O(l) + Na₃C₆H₅O₇(aq)

2. Find the limiting reagent between the reactants:

-For the $NaHCO_{3}$:

$1gNaHCO_{3}*\frac{1molNaHCO_{3}}{84gNaHCO_{3}}=0.012molesNaHCO_{3}$

Divide the number of moles between the stoichiometric coefficient of the $NaHCO_{3}$:

$\frac{0.012}{3}=0.004$

-For the $H_{3}C_{6}H_{5}O_{7}$:

$1gH_{3}C_{6}H_{5}O_{7}*\frac{1molH_{3}C_{6}H_{5}O_{7}}{192gH_{3}C_{6}H_{5}O_{7}}=0.005molesH_{3}C_{6}H_{5}O_{7}$

Divide the number of moles between the stoichiometric coefficient of the $H_{3}C_{6}H_{5}O_{7}$:

$\frac{0.005}{1}=0.005$

The smallest number is the 0.004 therefore, the limiting reagent is the sodium bicarbonate.

3. Calculate the grams of carbon dioxide:

$1gNaHCO_{3}*\frac{1molNaHCO_{3}}{94gNaHCO_{3}}*\frac{3molesCO_{2}}{3molesNaHCO_{3}}*\frac{44gCO_{2}}{1molCO_{2}}=0.47gCO_{2}$