8.3 Reaction Stoichiometry, Limiting Reactants & Percent Yield

8.3 Reaction Stoichiometry, Limiting Reactants & Percent Yield

Quick Summary: Stoichiometry is the quantitative arithmetic of chemical reactions, using balanced chemical equations to relate quantities of reactants and products. Stoichiometric coefficients represent molar ratios, providing the central bridge connecting mass, solution volumes, and gas volumes across phases. In non-equimolar mixtures, the limiting reactant is completely consumed first, defining the theoretical yield of products. Percent yield quantifies practical efficiency, reflecting reversible equilibria, side reactions, and mechanical isolation losses.

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1. Reaction Stoichiometry: The Molar Bridge

A balanced chemical equation operates on the molecular and molar level. Stoichiometric coefficients define proportional relationships among discrete particles and moles: $a\,A + b\,B \longrightarrow c\,C + d\,D$ The ratio $c/a$ relates substance $A$ to product $C$. Every stoichiometric calculation between different chemical species must pass through this central mole-to-mole bridge.

General Mass-to-Mass Pathway

To calculate the mass of product $C$ formed from a given mass of reactant $A$: $\text{Mass } A \text{ (g)} \xrightarrow{\div M_A} \text{Moles } A \xrightarrow{\times (c/a)} \text{Moles } C \xrightarrow{\times M_C} \text{Mass } C \text{ (g)}$ $\text{Mass } C = \text{Mass } A \times \left( \frac{1\text{ mol } A}{M_A} \right) \times \left( \frac{c\text{ mol } C}{a\text{ mol } A} \right) \times \left( \frac{M_C}{1\text{ mol } C} \right)$

Solution & Gas Stoichiometry

• Aqueous Solutions: Quantities are measured as solution volume and molarity ($M$, in $\text{mol/L}$): $n = M \times V \quad (\text{with } V \text{ in liters})$
• Gases at Standard Temperature and Pressure (STP): At $0^\circ\text{C}$ and $1.000\text{ atm}$, one mole of an ideal gas occupies the standard molar volume ($V_m = 22.414\text{ L/mol}$): $V = n \times 22.414\text{ L/mol}$
• Gases under Non-STP Conditions: The Ideal Gas Law connects volume, temperature, pressure, and molar quantity: $n = \frac{PV}{RT} \quad \text{or} \quad V = \frac{nRT}{P}$ where $R = 0.082057\text{ L}\cdot\text{atm}/(\text{mol}\cdot\text{K})$.

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2. Limiting Reactants & Excess Reagents

Reactants are rarely combined in exact stoichiometric proportions. One reactant is typically exhausted before the others:

• Limiting Reactant: The reactant completely consumed first. Its depletion terminates the reaction and determines the maximum theoretical yield of products.
• Excess Reactant: Any reactant present in an amount greater than required to react with the limiting reactant. A portion remains unreacted after reaction completion.

Identifying Limiting Reactants

1. Theoretical Product Yield Method: Calculate moles of product each reactant can generate. The reactant generating the smallest product yield is limiting, and that yield is the theoretical yield.
2. Mole-to-Coefficient Ratio Method: Divide initial moles ($n_i$) of each reactant by its stoichiometric coefficient ($a_i$): $R_i = n_i / a_i$. The smallest $R_i$ indicates the limiting reactant.

Calculating Unreacted Excess Reagents

1. Calculate moles of excess consumed: $n_{\text{excess, consumed}} = n_{\text{limiting}} \times (\text{coeff}_{\text{excess}} / \text{coeff}_{\text{limiting}})$.
2. Calculate moles remaining: $n_{\text{excess, remaining}} = n_{\text{excess, initial}} - n_{\text{excess, consumed}}$.
3. Convert to mass: $m_{\text{excess, remaining}} = n_{\text{excess, remaining}} \times M_{\text{excess}}$.

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3. Theoretical, Actual & Percent Yield

• Theoretical Yield: Maximum calculated mass of product formed if 100% of limiting reactant converts without loss.
• Actual Yield: Mass of isolated, purified product collected experimentally.
• Percent Yield: Efficiency metric defined as: $\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\%$

Factors Governing Yield Deviations

• Yields < 100%: Incomplete conversion due to dynamic chemical equilibrium; competing side reactions; mechanical losses during filtration, transfer, or recrystallization; loss of volatile compounds.
• Apparent Yields > 100%: Incomplete drying of isolated solids leaving residual solvent or water mass; co-precipitation of impurities.

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4. Comprehensive Worked Stoichiometry Problem

Titanium metal is produced via the Kroll process at 850 °C: $\text{TiCl}_4(g) + 2\,\text{Mg}(l) \longrightarrow \text{Ti}(s) + 2\,\text{MgCl}_2(l)$

Suppose $37.93\text{ g}$ of $\text{TiCl}_4$ ($M = 189.68\text{ g/mol}$) reacts with $12.16\text{ g}$ of $\text{Mg}$ ($M = 24.305\text{ g/mol}$).

1. Limiting Reactant Determination

• $n_{\text{TiCl}_4} = 37.93 / 189.68 = 0.2000\text{ mol}$
• $n_{\text{Mg}} = 12.16 / 24.305 = 0.5003\text{ mol}$
• Ratios: $R_{\text{TiCl}_4} = 0.2000 / 1 = 0.2000$; $R_{\text{Mg}} = 0.5003 / 2 = 0.2502$.
• Because $R_{\text{TiCl}_4} < R_{\text{Mg}}$, $\text{TiCl}_4$ is the limiting reactant, and $\text{Mg}$ is in excess.

2. Theoretical Yield of Titanium

• $n_{\text{Ti}} = 0.2000\text{ mol } \text{TiCl}_4 \times (1\text{ mol Ti} / 1\text{ mol } \text{TiCl}_4) = 0.2000\text{ mol Ti}$
• $m_{\text{Ti}} = 0.2000\text{ mol} \times 47.867\text{ g/mol} = 9.573\text{ g Ti}$

3. Mass of Excess Magnesium Remaining

• $n_{\text{Mg, consumed}} = 0.2000\text{ mol} \times 2 = 0.4000\text{ mol Mg}$
• $n_{\text{Mg, remaining}} = 0.5003 - 0.4000 = 0.1003\text{ mol Mg}$
• $m_{\text{Mg, remaining}} = 0.1003\text{ mol} \times 24.305\text{ g/mol} = 2.438\text{ g Mg}$

4. Percent Yield

If $8.240\text{ g}$ of purified titanium is collected: $\text{Percent Yield} = \left( \frac{8.240\text{ g}}{9.573\text{ g}} \right) \times 100\% = 86.08\%$