Question

X g of benzoic acid on reaction with aqueous \(NaHCO_3\) release \(CO_2\) that occupied 11.2 L volume at STP. X is ________ g.

Hint:

In such reactions, always use the stoichiometric relationship from the balanced chemical equation to convert volumes of gases into moles. Then use the molar mass to find the corresponding mass.

Answer 1

Step 1: Balanced Chemical Equation

The reaction between benzoic acid (C\(_6\)H\(_5\)COOH) and sodium bicarbonate (NaHCO\(_3\)) is represented by the following balanced equation: \[ \text{C}_6\text{H}_5\text{COOH(aq)} + \text{NaHCO}_3\text{(aq)} \rightarrow \text{C}_6\text{H}_5\text{COONa(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)} \]

Step 2: Moles of CO\(_2\) Released Calculation

At Standard Temperature and Pressure (STP), 1 mole of any gas occupies 22.4 L. Given that 11.2 L of CO\(_2\) was released, the moles of CO\(_2\) are calculated as: \[ n_{\text{CO}_2} = \frac{11.2 \, \text{L}}{22.4 \, \text{L/mol}} = 0.5 \, \text{mol} \]

Step 3: Moles of Benzoic Acid Calculation

According to the balanced chemical equation, 1 mole of benzoic acid yields 1 mole of CO\(_2\). Consequently, the moles of benzoic acid are equivalent to the moles of CO\(_2\): \[ n_{\text{Benzoic Acid}} = n_{\text{CO}_2} = 0.5 \, \text{mol} \]

Step 4: Mass of Benzoic Acid Calculation

The molar mass of benzoic acid (C\(_6\)H\(_5\)COOH) is determined as follows: \[ \text{Molar Mass} = 6(12) + 5(1) + 12 + 2(16) + 1 = 72 + 5 + 12 + 32 + 1 = 122 \, \text{g/mol} \] The mass of benzoic acid is then calculated using the formula: \[ X = n_{\text{Benzoic Acid}} \times \text{Molar Mass} = 0.5 \, \text{mol} \times 122 \, \text{g/mol} = 61 \, \text{g} \]

Conclusion

The mass of benzoic acid is \( \boxed{61} \) grams.