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: Write the balanced chemical equation.
Benzoic acid (C$_6$H$_5$COOH) reacts with NaHCO$_3$ as follows: C$_6$H$_5$COOH(aq) + NaHCO$_3$(aq) $\rightarrow$ C$_6$H$_5$COONa(aq) + H$_2$O(l) + CO$_2$(g)
Step 2: Calculate the moles of CO$_2$ released.
At STP, 1 mole of any gas occupies 22.4 L. Therefore, the number of moles of CO$_2$ released is:
$n_{{CO}_2} = \frac{11.2 { L}}{22.4 { L/mol}} = 0.5 { mol}$ 
Step 3: Calculate the moles of benzoic acid. From the balanced equation, 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$_2$:
$n_{{Benzoic Acid}} = n_{{CO}_2} = 0.5 { mol}$ 
Step 4: Calculate the mass of benzoic acid.
The molar mass of benzoic acid (C$_6$H$_5$COOH) is:
6(12) + 5(1) + 12 + 2(16) + 1 = 72 + 5 + 12 + 32 + 1 = 122 g/mol The mass of benzoic acid is:
$X = n_{{Benzoic Acid}} \times {Molar Mass} = 0.5 { mol} \times 122 { g/mol} = 61 { g}$
Therefore, X is 61 g.