To solve the given problem, we need to address the three questions step-by-step:

Step 1: Chemical Equation and Molar Mass

The balanced chemical equation is: $3 \text{NaHCO}_3 + \text{H}_3\text{C}_6\text{H}_5\text{O}_7 \rightarrow 3 \text{CO}_2 + 3 \text{H}_2\text{O} + \text{Na}_3\text{C}_6\text{H}_5\text{O}_7$

Molar masses (approximately):

• $\text{NaHCO}_3$: $84 \, \text{g/mol}$
• $\text{H}_3\text{C}_6\text{H}_5\text{O}_7$: $192 \, \text{g/mol}$
• $\text{CO}_2$: $44 \, \text{g/mol}$

Step 2: Determine the Limiting Reactant

Given masses:

• $300 \, \text{g}$ of $\text{NaHCO}_3$
• $300 \, \text{g}$ of $\text{H}_3\text{C}_6\text{H}_5\text{O}_7$

Convert to moles: $\text{Moles of } \text{NaHCO}_3 = \frac{300}{84} \approx 3.57 \, \text{mol}$ $\text{Moles of } \text{H}_3\text{C}_6\text{H}_5\text{O}_7 = \frac{300}{192} \approx 1.56 \, \text{mol}$

From the balanced equation, the reaction requires:

• 3 moles of $\text{NaHCO}_3$ for 1 mole of $\text{H}_3\text{C}_6\text{H}_5\text{O}_7$.

To find the limiting reactant, compare the actual mole ratio: $\text{Ratio of } \frac{\text{NaHCO}_3}{\text{H}_3\text{C}_6\text{H}_5\text{O}_7} = \frac{3.57}{1.56} \approx 2.29$ The reaction requires a ratio of 3:1, but we have less $\text{NaHCO}_3$ than needed. Therefore, $\text{NaHCO}_3$ is the limiting reactant.

Step 3: Mass of $\text{CO}_2$ Formed

From the balanced equation:

• 3 moles of $\text{NaHCO}_3$ produce 3 moles of $\text{CO}_2$.

Moles of $\text{CO}_2$ produced: $\text{Moles of } \text{CO}_2 = \text{Moles of } \text{NaHCO}_3 = 3.57 \, \text{mol}$

Mass of $\text{CO}_2$: $\text{Mass of } \text{CO}_2 = 3.57 \times 44 \approx 157.08 \, \text{g}$

Step 4: Mass of Excess Reactant Remaining

Excess reactant: $\text{H}_3\text{C}_6\text{H}_5\text{O}_7$.

Moles of $\text{H}_3\text{C}_6\text{H}_5\text{O}_7$ consumed: $\text{Moles consumed} = \frac{\text{Moles of } \text{NaHCO}_3}{3} = \frac{3.57}{3} \approx 1.19 \, \text{mol}$

Moles of $\text{H}_3\text{C}_6\text{H}_5\text{O}_7$ remaining: $\text{Moles remaining} = 1.56 - 1.19 = 0.37 \, \text{mol}$

Mass of $\text{H}_3\text{C}_6\text{H}_5\text{O}_7$ remaining: $\text{Mass remaining} = 0.37 \times 192 \approx 71.04 \, \text{g}$

Final Answers:

1. Limiting reactant: $\text{NaHCO}_3$.
2. Mass of $\text{CO}_2$ formed: $157.08 \, \text{g}$.
3. Mass of excess reactant remaining: $71.04 \, \text{g}$.

Would you like detailed step-by-step explanations for any of these calculations?

Expanding Questions:

1. How do stoichiometric coefficients in a reaction affect limiting reactant determination?
2. Why is it important to identify the limiting reactant in chemical reactions?
3. How would the results differ if $\text{H}_3\text{C}_6\text{H}_5\text{O}_7$ was the limiting reactant instead?
4. What are practical applications of this type of stoichiometric analysis?
5. Can the reaction yield be affected by factors other than the limiting reactant, such as reaction conditions?

Tip: Always double-check molar masses and stoichiometric ratios when performing chemical calculations to avoid errors.