Alcohols, Phenols and Ethers: Numerical Problems Set (7)

Question

When $4.6\,\text{g}$ of ethanol is treated with excess sodium metal, the volume of $\text{H}_2$ gas collected at STP is what?

Solution — Step by Step

The volume of $\text{H}_2$ at STP is $1.12\,\text{L}$ (or $1120\,\text{mL}$).

Why This Works

Sodium reacts with the OH group of alcohols (and phenols) liberating exactly half a mole of $\text{H}_2$ per mole of OH. Counting the OH groups is the trick — diols give twice as much $\text{H}_2$ as monohydric alcohols of the same molar quantity.

The stoichiometry $2\,\text{ROH} \to \text{H}_2$ is key. One H from each ROH combines to form $\text{H}_2$, so two ROHs produce one $\text{H}_2$.

Alternative Method

For sodium-with-OH problems, use the shortcut: $\text{moles H}_2 = \text{moles OH}/2$. With $0.1\,\text{mol}$ ethanol (one OH each), $0.05\,\text{mol H}_2$, $1.12\,\text{L}$ at STP.

Common Mistake

Forgetting the factor of $2$ in the stoichiometry. Students compute $n_{\text{H}_2} = n_{\text{ROH}}$ and double the answer. Always balance the equation before substituting numbers.