Question
A sample of crystalline soda A with a mass of 1.287 g was allowed to react with an excess of hydrochloric acid and 100.8 mL of a gas was liberated (measured at STP). How many molecules of water in relation to one molecule of ${Na}_{2}{CO}_{3}$ are contained in the sample of soda?
(Relative atomic masses are: $Na = 23;H = 1;C = 12;O = 16$ )
Options
Solution
Sol. Key concept: Moles of carbon dioxide determine moles of hydrated sodium carbonate.
Write the crystalline soda as Na₂CO₃·xH₂O. One mole of hydrate contains one carbonate group and produces one mole of CO₂ with excess acid.
$$Na_{2}CO_{3} \cdot \ xH_{2}O + 2HCl \rightarrow 2NaCl + CO_{2} + (x + 1)H_{2}O$$
Correction / clarification: The balanced equation contains (x + 1)H₂O on the right: x waters of crystallisation plus one water formed by carbonate neutralisation. The previous xH₂O product was unbalanced.
$$n(CO_{2}) = \frac{0.1008}{22.4} = 0.0045\ mol$$
$$M(hydrate) = \frac{1.287}{0.0045} = 286\ g\ mol^{- 1}$$
$$106 + 18x = 286\ \ \Rightarrow \ \ \ x = 10$$
Therefore, the salt is Na₂CO₃·10H₂O and the answer is (d).
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