Li3N(s) + 2 H2(g) <-> LiNH2(s) + 2 LiH(s) dH° = -192 kJ/mol_rxn Because pure H2 is a hazardous subst — Stoichiometry Chemistry Question

STEPS:

1. Calculate the moles of the limiting reactant, $\text{Li}_3\text{N}$:
* The problem states that $70.\text{ g}$ of solid lithium nitride ($\text{Li}_3\text{N}$) is reacted with excess hydrogen gas ($\text{H}_2$), making $\text{Li}_3\text{N}$ the limiting reactant.
* The molar mass of $\text{Li}_3\text{N}$ is given as $35\text{ g/mol}$.
* Calculate the initial moles of $\text{Li}_3\text{N}$ present:
$$\text{Moles of }\text{Li}_3\text{N} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{70.\text{ g}}{35\text{ g/mol}} = \mathbf{2.0\text{ mol}}$$

2. Use stoichiometry to determine the theoretical moles of $\text{LiH}$ produced:
* Refer to the balanced chemical equation:
$$\text{Li}_3\text{N}(s) + 2\ \text{H}_2(g) \rightleftharpoons \text{LiNH}_2(s) + 2\ \text{LiH}(s)$$
* According to the coefficients, $1\text{ mole}$ of $\text{Li}_3\text{N}$ reacts to produce $2\text{ moles}$ of $\text{LiH}$.
* Therefore, calculate the theoretical moles of $\text{LiH}$ that should be formed:
$$\text{Theoretical Moles of }\text{LiH} = 2.0\text{ mol }\text{Li}_3\text{N} \times \frac{2\text{ mol }\text{LiH}}{1\text{ mol }\text{Li}_3\text{N}} = \mathbf{4.0\text{ mol}}$$

3. Convert theoretical moles of $\text{LiH}$ to theoretical mass in grams:
* Determine the molar mass of lithium hydride ($\text{LiH}$) using atomic masses from the periodic table:
* Atomic mass of Lithium ($\text{Li}$) $\approx 6.94\text{ g/mol}$ (roughly $7\text{ g/mol}$).
* Atomic mass of Hydrogen ($\text{H}$) $\approx 1.01\text{ g/mol}$ (roughly $1\text{ g/mol}$).
* Molar mass of $\text{LiH} \approx 6.94 + 1.01 \approx \mathbf{8.0\text{ g/mol}}$.
* Calculate the theoretical yield in grams:
$$\text{Theoretical Mass of }\text{LiH} = 4.0\text{ mol} \times 8.0\text{ g/mol} = \mathbf{32\text{ g}}$$

4. Calculate the percent yield:
* The actual experimental mass of $\text{LiH}$ produced is $8.0\text{ g}$.
* Calculate the percent yield:
$$\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$$
$$\text{Percent Yield} = \frac{8.0\text{ g}}{32\text{ g}} \times 100\% = 0.25 \times 100\% = \mathbf{25\%}$$
* This identifies Option B as the correct choice.

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WHY_OTHERS_WRONG:

• Option A is incorrect (17%): A student might arrive at this value by using a $1:1$ mole ratio between $\text{Li}_3\text{N}$ and $\text{LiH}$ (ignoring the stoichiometric coefficient of 2), which would yield $2.0\text{ moles}$ of $\text{LiH}$ theoretically. If they also miscalculate the molar mass of $\text{LiH}$ (e.g., as $\approx 24\text{ g/mol}$), they would obtain a theoretical yield of $48\text{ g}$, leading to $\frac{8.0\text{ g}}{48\text{ g}} \approx 17\%$.
• Option C is incorrect (50.%): A student would calculate this if they forgot to apply the $1:2$ stoichiometric ratio between $\text{Li}_3\text{N}$ and $\text{LiH}$. Assuming a $1:1$ ratio, the theoretical mass is $2.0\text{ mol} \times 8.0\text{ g/mol} = 16\text{ g}$. This yields a percent yield of $\frac{8.0\text{ g}}{16\text{ g}} = 50\%$, which is exactly double the correct answer.
• Option D is incorrect (100%): This represents a perfect theoretical reaction where the entire limiting reactant is successfully converted without any losses, side reactions, or equilibrium limitations. This is physically contradicted by the experimental recovery of only $8.0\text{ g}$ of $\text{LiH}$ instead of the theoretical $32\text{ g}$.