To estimate the amount of citric acid ([tex]\(C_6H_8O_7\)[/tex]) in the tablet, you need to use stoichiometry and the balanced chemical equation for the reaction between sodium bicarbonate ([tex]\(NaHCO_3\)[/tex]) and citric acid. Here’s the step-by-step explanation:
1. Identify the Balanced Chemical Equation:
The reaction between citric acid and sodium bicarbonate is:
[tex]\[
C_6H_8O_7 + 3 \, NaHCO_3 \rightarrow 3 \, CO_2 + 3 \, H_2O + Na_3C_6H_5O_7
\][/tex]
According to the equation, one mole of citric acid reacts with three moles of sodium bicarbonate.
2. Determine the Amount of Sodium Bicarbonate:
From your data (referred to as on slide 11), you know the mass of sodium bicarbonate ([tex]\(NaHCO_3\)[/tex]) that has dissolved. Let's denote this mass as [tex]\( \text{mass_{NaHCO_3}} \)[/tex] in grams.
3. Convert the Mass of Sodium Bicarbonate to Moles:
Use the molar mass of sodium bicarbonate to convert its mass to moles. The molar mass of [tex]\(NaHCO_3\)[/tex] is approximately 84 g/mol.
[tex]\[
\text{moles_{NaHCO_3}} = \frac{\text{mass_{NaHCO_3}}}{\text{molar mass_{NaHCO_3}}}
\][/tex]
4. Use the Stoichiometric Ratio to Determine Moles of Citric Acid:
According to the balanced chemical equation, one mole of [tex]\(C_6H_8O_7\)[/tex] reacts with three moles of [tex]\(NaHCO_3\)[/tex]. Therefore, you divide the moles of [tex]\(NaHCO_3\)[/tex] by 3 to get the moles of citric acid.
[tex]\[
\text{moles_{C_6H_8O_7}} = \frac{\text{moles_{NaHCO_3}}}{3}
\][/tex]
5. Convert the Moles of Citric Acid to Mass:
Finally, convert the moles of citric acid back to mass. The molar mass of citric acid ([tex]\(C_6H_8O_7\)[/tex]) is approximately 192 g/mol.
[tex]\[
\text{mass_{C_6H_8O_7}} = \text{moles_{C_6H_8O_7}} \times \text{molar mass_{C_6H_8O_7}}
\][/tex]
6. Conclusion:
The computed mass, [tex]\(\text{mass_{C_6H_8O_7}}\)[/tex], represents the estimated amount of citric acid ([tex]\(C_6H_8O_7\)[/tex]) present in the tablet.
Using these steps, you can estimate the amount of citric acid in the tablet based on the amount of dissolved sodium bicarbonate.
