Question

An undergraduate student weighed 20g of sodium carbonate salt, if Na=23, O =16, carbon =12 and H=1, what is the mole of the hydroxide?

Answer 1

Answer on Question #55578 – Chemistry – Inorganic Chemistry

Questions:

An undergraduate student weighed 20g of sodium carbonate salt, if Na=23, O=16, carbon=12 and H=1, what is the mole of the hydroxide?

Solution:

$$\begin{array}{l} m(\mathrm{NaHCO_3}) = 20 \text{ g}; \\ Ar(\mathrm{Na}) = 23 \text{ g} \times \mathrm{mol^{-1}}; \\ Ar(\mathrm{O}) = 16 \text{ g} \times \mathrm{mol^{-1}}; \\ Ar(\mathrm{C}) = 12 \text{ g} \times \mathrm{mol^{-1}}; \\ Ar(\mathrm{H}) = 1 \text{ g} \times \mathrm{mol^{-1}}; \\ n(\mathrm{OH^-}) - ? \\ \end{array}$$ $$\begin{array}{l} M(\mathrm{NaHCO_3}) = Ar(\mathrm{Na}) + Ar(\mathrm{H}) + Ar(\mathrm{C}) + 3Ar(\mathrm{O}); \\ M(\mathrm{NaHCO_3}) = 84 \text{ g} \times \mathrm{mol^{-1}}; \\ n = \frac{m}{M}; \\ n(\mathrm{NaHCO_3}) = \frac{m(\mathrm{NaHCO_3})}{M(\mathrm{NaHCO_3})}; \\ n(\mathrm{NaHCO_3}) = 0.238 \text{ mol}; \\ \end{array}$$

One molecule of NaHCO₃ contains one OH⁻ group. According to this statement:

$$n(\mathrm{OH^-}) = n(\mathrm{NaHCO_3}) = 0.238 \text{ mol};$$

Answer: 0.238 mol

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