A batch of dry ice (solid CO₂) was contaminated during manufacture - HSC - SSCE Chemistry - Question 30 - 2060 - Paper 1

First, determine the moles of NaOH that reacted with CO₂:

From the reaction:

$2NaOH + CO₂ \rightarrow Na₂CO₃ + H₂O$

This means that 2 moles of NaOH react with 1 mole of CO₂. The total moles of NaOH used in neutralizing CO₂ can be calculated by rearranging the above equation:

Let 0.0500 mol be the initial moles of NaOH added, and let the moles of NaOH left after the titration be calculated from the HCl used.

Using HCl to titrate remaining NaOH:

$n(HCl) = cV = 0.0276 \, \text{mol L}^{-1} \times 0.02760 \, \text{L} = 0.00076376 \, \text{mol}$

This means:

$n(NaOH reacted) = 0.0500 \, \text{mol} - 0.00076376 \, \text{mol} = 0.04923624 \, \text{mol}$

From the stoichiometry:

$n(CO₂) = \frac{1}{2} n(NaOH) = \frac{1}{2} \times 0.04923624 \, \text{mol} = 0.02461812 \, \text{mol}$

Now, calculating the mass of CO₂:

$\text{Mass}(CO₂) = n \times M = 0.02461812 \, \text{mol} \times 44.01 \, \text{g mol}^{-1} = 1.082 \, \text{g}$

Lastly, the percentage purity by mass of the batch of dry ice is:

$\% \text{Pure} = \frac{\text{Mass of CO₂}}{\text{Total mass of dry ice}} \times 100\% = \frac{1.082 \, \text{g}}{0.616 \, \text{g}} \times 100\% = 175.0\%$

This should be calculated correctly and checked as the direct fraction must be less than 100%. An error should be looked for in volumes and calculation.