Some CO\(_2\) gas was kept in a sealed container at a pressure of 1 atm and at 273 K. This entire amount of CO\(_2\) gas was later passed through an aqueous solution of Ca(OH)\(_2\). The excess unreacted Ca(OH)\(_2\) was later neutralized with 0.1 M of 40 mL HCl. If the volume of the sealed container of CO\(_2\) was \(x\), then \(x\) is ________________________ cm\(^3\) (nearest integer).
[Given: The entire amount of CO\(_2\) reacted with exactly half the initial amount of Ca(OH)\(_2\) present in the aqueous solution.]
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When dealing with gas volume at standard conditions, use the molar volume of gas at STP, which is 22400 cm³ per mole.
Let \(n\) be the initial moles of CO\(_2\) and \(2n\) be the initial moles of Ca(OH)\(_2\). The excess Ca(OH)\(_2\) is \(n\) moles. The gram equivalent of Ca(OH)\(_2\) equals the gram equivalent of HCl. Using the molarity and volume of HCl, we have:\[n \times 2 = 0.1 \times \frac{40}{1000}\]\[n = 2 \times 10^{-3}\]The volume of CO\(_2\) is calculated as:\[2 \times 10^{-3} \times 22400 = 44.8 \, \text{cm}^3\]Therefore, the volume of CO\(_2\) is approximately 45 cm\(^3\).
