A student added 627 mg of hydrated sodium carbonate (Na₂CO₃·xH₂O) to 200 cm³ of 0.250 mol dm⁻³ hydrochloric acid in a beaker and stirred the mixture - AQA - A-Level Chemistry - Question 10 - 2018 - Paper 1

To find the moles of NaOH used to neutralize the remaining HCl after the reaction with Na₂CO₃·xH₂O, we can use the mean titre value of 26.60 cm³, converted to dm³:

$ext{Volume of NaOH} = 26.60 ext{ cm}^{3} = 0.0266 ext{ dm}^{3}$

The moles of NaOH used is then:

$ext{Moles of NaOH} = ext{Concentration} imes ext{Volume} = 0.150 ext{ mol dm}^{-3} imes 0.0266 ext{ dm}^{3} = 0.00399 ext{ mol}$

The moles of HCl that reacted with Na₂CO₃·xH₂O is:

$ext{Moles of HCl reacted} = ext{Initial moles of HCl} - ext{Moles of NaOH} = 0.050 - 0.00399 = 0.04601 ext{ mol}$

From the reaction stoichiometry:

ightarrow 2 ext{NaCl} + ext{CO₂} + ext{H₂O}$$ 1 mole of Na₂CO₃·xH₂O reacts with 2 moles of HCl. Therefore: $$5x = rac{0.04601}{2} = 0.023005 ext{ mol}$$

To calculate the moles of Na₂CO₃·xH₂O used:

$ext{Mass of Na₂CO₃·xH₂O} = 627 ext{ mg} = 0.627 ext{ g}$

Molar mass of Na₂CO₃ is approximately 106 g/mol. This gives:

ext{Moles of Na₂CO₃·xH₂O} = rac{0.627}{106} = 0.00591 ext{ mol}

Combining the previous results:

$5x = 0.00591 ext{ mol}$

Thus,

x = rac{0.00591}{5} = 0.001182 ext{ mol}

Since each mole corresponds to 18 grams of water:

$ext{Value of } x = 1$