Answer:
0.7 grams of CO₂ is produced during the reaction between citric acid (C₆H₈O₇) and sodium bicarbonate (NaHCO₃).
General Formulas and Concepts:
Atomic Structure and Properties
Chemical Reactions
Explanation:
Step 1: Define
Identify given.
[RxN]: C₆H₈O₇ + 3NaHCO₃ → 3H₂O + 3CO₂ + Na₃C₆H₅O₇
[Limiting Reactant]: 1 g C₆H₈O₇
Step 2: Find CO₂ Amount
We know that our reaction is in terms of moles. In order to find the amount of carbon dioxide released, we will need to convert our units into moles.
To start out, we are given 1 g citric acid, C₆H₈O₇, which will be completely used up. From the Periodic Table (see attachment), we can determine our molar mass of citric acid for mole conversion:
- Molar Mass of C: 12.01 g/mol
- Molar Mass of H: 1.01 g/mol
- Molar Mass of O: 16.00 g/mol
Combining these, we can find out molar mass of citric acid:
- Molar Mass of C₆H₈O₇: 6(12.01 g/mol) + 8(1.01 g/mol) + 7(16.00 g/mol) = 192.14 g/mol
We also need to find out the molar mass of CO₂:
- Molar Mass of CO₂: 12.01 g/mol + 2(16.00 g/mol) = 44.01 g/mol
Now we can set up our dimensional analysis to find the amount of CO₂ released:
[tex]\displaystyle 1 \ \text{g} \ \text{C}_6 \text{H}_8 \text{O}_7 \bigg( \frac{1 \ \text{mol} \ \text{C}_6 \text{H}_8 \text{O}_7}{192.14 \ \text{g} \ \text{C}_6 \text{H}_8 \text{O}_7} \bigg) \bigg( \frac{3 \ \text{mol} \ \text{C} \text{O}_2}{1 \ \text{mol} \ \text{C}_6 \text{H}_8 \text{O}_7} \bigg) \bigg( \frac{44.01 \ \text{g} \ \text{C} \text{O}_2}{1 \ \text{mol} \ \text{C} \text{O}_2} \bigg)[/tex]
We can see that all units cross out and what we are left with is grams of CO₂, which is what we are trying to find.
Evaluating our dimensional analysis, we obtain an answer:
[tex]\displaystyle 1 \ \text{g} \ \text{C}_6 \text{H}_8 \text{O}_7 \bigg( \frac{1 \ \text{mol} \ \text{C}_6 \text{H}_8 \text{O}_7}{192.14 \ \text{g} \ \text{C}_6 \text{H}_8 \text{O}_7} \bigg) \bigg( \frac{3 \ \text{mol} \ \text{C} \text{O}_2}{1 \ \text{mol} \ \text{C}_6 \text{H}_8 \text{O}_7} \bigg) \bigg( \frac{44.01 \ \text{g} \ \text{C} \text{O}_2}{1 \ \text{mol} \ \text{C} \text{O}_2} \bigg) = 0.687155 \ \text{g} \ \text{C} \text{O}_2[/tex]
From our given problem, we are given 1 significant figure. Let us round our answer to obtain our final answer:
[tex]\displaystyle \begin{aligned}\text{Amount of carbon dioxide produced} & = 0.687155 \ \text{g} \ \text{C} \text{O}_2 \\& \approx \boxed{ 0.7 \ \text{g} \ \text{C} \text{O}_2 }\end{aligned}[/tex]
∴ 1 gram of citric acid produces 0.7 grams of carbon dioxide during the reaction.
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Topic: AP Chemistry
Unit: Chemical Reactions
