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.
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.
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Approach Solution - 1
Step 1: Write the Balanced Chemical Equation
The reaction between benzoic acid (C\(_6\)H\(_5\)COOH) and sodium bicarbonate (NaHCO\(_3\)) is: \[ \text{C}_6\text{H}_5\text{COOH(aq)} + \text{NaHCO}_3\text{(aq)} \rightarrow \text{C}_6\text{H}_5\text{COONa(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)} \]
Step 2: Calculate the Moles of CO\(_2\) Released
At STP (Standard Temperature and Pressure), 1 mole of any gas occupies 22.4 L. Given that 11.2 L of CO\(_2\) is released, the number of moles of CO\(_2\) is: \[ n_{\text{CO}_2} = \frac{11.2 \, \text{L}}{22.4 \, \text{L/mol}} = 0.5 \, \text{mol} \]
Step 3: Calculate the Moles of Benzoic Acid
From the balanced chemical equation, we can see that 1 mole of benzoic acid reacts to produce 1 mole of CO\(_2\). Therefore, the number of moles of benzoic acid is equal to the number of moles of CO\(_2\): \[ n_{\text{Benzoic Acid}} = n_{\text{CO}_2} = 0.5 \, \text{mol} \]
Step 4: Calculate the Mass of Benzoic Acid
The molar mass of benzoic acid (C\(_6\)H\(_5\)COOH) is: \[ \text{Molar Mass} = 6(12) + 5(1) + 12 + 2(16) + 1 = 72 + 5 + 12 + 32 + 1 = 122 \, \text{g/mol} \] The mass of benzoic acid is calculated as: \[ X = n_{\text{Benzoic Acid}} \times \text{Molar Mass} = 0.5 \, \text{mol} \times 122 \, \text{g/mol} = 61 \, \text{g} \]
Conclusion
The mass of benzoic acid is \( \boxed{61} \) grams.
Approach Solution -2
Given: In the reaction of benzoic acid with aqueous sodium bicarbonate (\( \text{NaHCO}_3 \)), CO₂ is released. The volume of CO₂ produced is 11.2 L at STP. We know that at STP (Standard Temperature and Pressure), 1 mole of gas occupies 22.4 L. The balanced chemical equation for the reaction is: \[ \text{C}_6\text{H}_5\text{COOH (aq)} + \text{NaHCO}_3 \rightarrow \text{C}_6\text{H}_5\text{COONa (aq)} + \text{CO}_2 (g) + \text{H}_2\text{O (l)}. \]
Step 1: Moles of CO₂ Released
The volume of CO₂ released is given as 11.2 L. Using the molar volume of gas at STP (22.4 L = 1 mole), we can calculate the number of moles of CO₂: \[ \text{Moles of CO}_2 = \frac{11.2 \, \text{L}}{22.4 \, \text{L/mol}} = 0.5 \, \text{moles of CO}_2. \]
Step 2: Moles of Benzoic Acid Used
From the balanced equation, 1 mole of benzoic acid reacts with 1 mole of sodium bicarbonate to produce 1 mole of CO₂. Therefore, the moles of benzoic acid used will be the same as the moles of CO₂ produced: \[ \text{Moles of benzoic acid} = 0.5 \, \text{moles}. \]
Step 3: Mass of Benzoic Acid Used
The molar mass of benzoic acid (\( \text{C}_6\text{H}_5\text{COOH} \)) is 122 g/mol. Thus, the mass of benzoic acid used is: \[ \text{Mass of benzoic acid} = \text{moles} \times \text{molar mass} = 0.5 \, \text{moles} \times 122 \, \text{g/mol} = 61 \, \text{g}. \]
Final Answer:
The mass of benzoic acid used is \( \boxed{61} \, \text{g}. \)
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