Answer:
0.47g of [tex]CO_{2}[/tex]
Explanation:
1. Balanced equation:
3NaHCO₃(aq) + H₃C₆H₅O₇(aq) --------> 3CO₂(g) + 3H₂O(l) + Na₃C₆H₅O₇(aq)
2. Find the limiting reagent between the reactants:
-For the [tex]NaHCO_{3}[/tex]:
[tex]1gNaHCO_{3}*\frac{1molNaHCO_{3}}{84gNaHCO_{3}}=0.012molesNaHCO_{3}[/tex]
Divide the number of moles between the stoichiometric coefficient of the [tex]NaHCO_{3}[/tex]:
[tex]\frac{0.012}{3}=0.004[/tex]
-For the [tex]H_{3}C_{6}H_{5}O_{7}[/tex]:
[tex]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}[/tex]
Divide the number of moles between the stoichiometric coefficient of the [tex]H_{3}C_{6}H_{5}O_{7}[/tex]:
[tex]\frac{0.005}{1}=0.005[/tex]
The smallest number is the 0.004 therefore, the limiting reagent is the sodium bicarbonate.
3. Calculate the grams of carbon dioxide:
[tex]1gNaHCO_{3}*\frac{1molNaHCO_{3}}{94gNaHCO_{3}}*\frac{3molesCO_{2}}{3molesNaHCO_{3}}*\frac{44gCO_{2}}{1molCO_{2}}=0.47gCO_{2}[/tex]
