A 0.35 g sample of Li(s) is placed in an Erlenmeyer flask containing 100 mL of water at 25°C. A ball — Stoichiometry Chemistry Question

STEPS:

1. Write down the balanced chemical equation representing the reaction of an alkali metal ($\text{M}$) with water: Both lithium ($\text{Li}$) and potassium ($\text{K}$) are Group 1 alkali metals and react with water in an identical stoichiometric ratio:
$$2\ \text{M}(s) + 2\ \text{H}_2\text{O}(l) \rightarrow 2\ \text{M}^+(aq) + 2\ \text{OH}^-(aq) + \text{H}_2(g)$$
This equation shows that for every $2\text{ moles}$ of alkali metal consumed, $1\text{ mole}$ of $\text{H}_2(g)$ is produced.
2. Recall the relationship between mass, molar mass, and moles: The number of moles ($n$) of a substance is determined by dividing its mass ($m$) by its molar mass ($\mathcal{M}$):
$$n = \frac{m}{\mathcal{M}}$$
3. Compare the molar masses of Lithium and Potassium using the Periodic Table:
* Molar mass of $\text{Li}$: $\approx 6.94\text{ g/mol}$
* Molar mass of $\text{K}$: $\approx 39.10\text{ g/mol}$
4. Calculate and compare the moles of metal present in each $0.35\text{ g}$ sample:
* Moles of $\text{Li}$ in the first experiment:
$$n_{\text{Li}} = \frac{0.35\text{ g}}{6.94\text{ g/mol}} \approx \mathbf{0.050\text{ mol}}$$
* Moles of $\text{K}$ in the second experiment:
$$n_{\text{K}} = \frac{0.35\text{ g}}{39.10\text{ g/mol}} \approx \mathbf{0.0090\text{ mol}}$$
* Because potassium has a much larger molar mass than lithium, $0.35\text{ g}$ of potassium contains significantly fewer moles of metal atoms than $0.35\text{ g}$ of lithium.
5. Relate the moles of reactant to the moles of gas produced:
* Since both metals react with the same stoichiometry, the reaction with fewer moles of metal reactant will yield fewer moles of gaseous $\text{H}_2$ product:
* $\text{Moles of }\text{H}_2\text{ from Li} = \frac{1}{2} \times 0.050\text{ mol} = \mathbf{0.025\text{ mol}}$
* $\text{Moles of }\text{H}_2\text{ from K} = \frac{1}{2} \times 0.0090\text{ mol} = \mathbf{0.0045\text{ mol}}$
6. Conclude the overall effect on gas production: Because the $0.35\text{ g}$ sample of potassium contains far fewer moles of reactant than the $0.35\text{ g}$ sample of lithium, a much smaller amount of hydrogen gas will be produced when potassium is used, which corresponds to Option B.

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WHY_OTHERS_WRONG:

• Option A is incorrect: Potassium is a highly reactive Group 1 alkali metal (even more reactive than lithium due to its lower first ionization energy). It reacts vigorously with water to generate hydrogen gas, so some gas will definitely be produced.
• Option C is incorrect: Equal quantities of gas would only be produced if the two experiments started with equal *moles* of the metals rather than equal *masses*. Because potassium atoms are much heavier than lithium atoms, a $0.35\text{ g}$ sample of potassium contains far fewer atoms (moles) of reactant.
• Option D is incorrect: While potassium is more reactive (kinetically faster) than lithium, the total *amount* (yield) of product is determined solely by stoichiometry and the moles of the limiting reactant. Because we have far fewer moles of potassium, it yields less total gas product than lithium.