9.2 g of ethanol is treated with an excess of sodium metal until the reaction is complete. What mass of sodium ethoxide is formed? (Atomic masses: H = 1, C = 12, O = 16, Na = 23)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 13.6 g
Alcohols react with active metals such as sodium to give the alkoxide and hydrogen. Each molecule of ethanol gives one formula unit of sodium ethoxide; two molecules together give one molecule of hydrogen:
\( \displaystyle \ce{2C2H5OH + 2Na -> 2C2H5ONa + H2} \)
\( \displaystyle n(\ce{C2H5OH}) = \frac{9.2\ \mathrm{g}}{46\ \mathrm{g\ mol^{-1}}} = 0.200\ \mathrm{mol} = n(\ce{C2H5ONa}) \)
\( \displaystyle M(\ce{C2H5ONa}) = 2 \times 12 + 5 \times 1 + 16 + 23 = 68\ \mathrm{g\ mol^{-1}} \)
\( \displaystyle m(\ce{C2H5ONa}) = 0.200\ \mathrm{mol} \times 68\ \mathrm{g\ mol^{-1}} = 13.6\ \mathrm{g} \)