# 4.2 Limiting and Non-Limiting Reactants In a chemical reaction, reactants are often not mixed in the exact stoichiometric ratios required by the balanced chemical equation. This means one reactant will be completely used up before the others. The **limiting reactant** (or limiting reagent) is the reactant that is completely consumed first in a chemical reaction. The amount of product formed is determined by this reactant. The **non-limiting reactant** (or excess reactant) is the reactant that is present in a quantity greater than what is required to react completely with the limiting reactant. Some of this reactant will be left over after the reaction is complete. ### Identifying the Limiting Reactant To understand this concept, consider the synthesis of water from hydrogen and oxygen: $$ 2H_2(g) + O_2(g) \rightarrow 2H_2O(l) $$ This balanced equation tells us that 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water. * **Scenario:** Suppose we mix 1 mole of $H_2$ and 1 mole of $O_2$. * According to the stoichiometry (2:1 ratio), 1 mole of $H_2$ requires $1/2$ mole of $O_2$ to react completely. * We have 1 mole of $O_2$ available, which is more than the required $1/2$ mole. Therefore, $O_2$ is in excess. * Conversely, for the entire 1 mole of $O_2$ to react, it would need 2 moles of $H_2$. We only have 1 mole of $H_2$. * Since the amount of $H_2$ is insufficient to react with all the $O_2$, **$H_2$ is the limiting reactant**. It will be completely consumed and will "limit" the amount of water produced. * **$O_2$ is the non-limiting or excess reactant.** The general procedure to identify the limiting reactant is: 1. Calculate the number of moles of each reactant present. 2. Use the stoichiometry of the balanced equation to calculate the amount of product that could be formed from each reactant. 3. The reactant that produces the *least* amount of product is the limiting reactant. ### Example 4.11: Synthesis of Calcium Nitride Calculate the mass of calcium nitride ($Ca_3N_2$) prepared from 54.9 grams of Ca and 43.2 grams of $N_2$. The balanced chemical equation is: $$ 3Ca(s) + N_2(g) \rightarrow Ca_3N_2(s) $$ **Solution:** **1. Calculate the moles of each reactant.** * Molar mass of Calcium (Ca) = $40.0 \, g/mol$ * Molar mass of Nitrogen gas ($N_2$) = $2 \times 14.0 = 28.0 \, g/mol$ $$ \text{Moles of Ca} = \frac{\text{mass}}{\text{molar mass}} = \frac{54.9 \, g}{40.0 \, g/mol} = 1.37 \, \text{moles} $$ $$ \text{Moles of } N_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{43.2 \, g}{28.0 \, g/mol} = 1.54 \, \text{moles} $$ **2. Determine the amount of product formed from each reactant.** * **From Ca:** The ratio of $Ca$ to $Ca_3N_2$ is 3:1. $$ 1.37 \, \text{moles Ca} \times \left( \frac{1 \, \text{mole } Ca_3N_2}{3 \, \text{moles Ca}} \right) = 0.457 \, \text{moles of } Ca_3N_2 $$ * **From $N_2$**: The ratio of $N_2$ to $Ca_3N_2$ is 1:1. $$ 1.54 \, \text{moles } N_2 \times \left( \frac{1 \, \text{mole } Ca_3N_2}{1 \, \text{mole } N_2} \right) = 1.54 \, \text{moles of } Ca_3N_2 $$ **3. Identify the limiting reactant.** * Calcium (Ca) produces the least amount of product (0.457 moles). * Therefore, **Ca is the limiting reactant**. **4. Calculate the mass of the product formed.** The maximum amount of product that can be formed is determined by the limiting reactant. * Molar mass of $Ca_3N_2 = 3(40.0) + 2(14.0) = 120.0 + 28.0 = 148.0 \, g/mol$ $$ \text{Mass of } Ca_3N_2 = \text{moles} \times \text{molar mass} = 0.457 \, \text{mol} \times 148.0 \, g/mol = \mathbf{67.64 \, g} $$ ### Possible Questions/Answers (Concept Assessment Exercise 4.5) **1. Q: How much volume is occupied by marsh gas ($CH_4$) at STP containing $4.8 \times 10^{24}$ molecules?** **A:** 1. **Calculate moles:** $$ \text{moles} = \frac{\text{Number of molecules}}{\text{Avogadro's Number}} = \frac{4.8 \times 10^{24}}{6.022 \times 10^{23} \, mol^{-1}} \approx 7.97 \, \text{mol} $$ 2. **Calculate volume at STP:** At STP, 1 mole of any ideal gas occupies 22.4 dm³. $$ \text{Volume} = \text{moles} \times \text{Molar Volume} = 7.97 \, \text{mol} \times 22.4 \, dm^3/mol \approx \mathbf{178.5 \, dm^3} $$ **2. Q: Calculate the molecular weight of a gas with a density of $1.25 \, g/dm^3$ at STP.** **A:** 1. At STP, we know the molar volume is $22.4 \, dm^3/mol$. 2. Molecular Weight (Molar Mass) can be found by: $$ M = \text{density} \times \text{Molar Volume} $$ $$ M = 1.25 \, g/dm^3 \times 22.4 \, dm^3/mol = \mathbf{28.0 \, g/mol} $$ *(This is the molar mass of nitrogen gas, $N_2$)* **3. Q: Calculate molecules of $H_2O$ produced by 500 molecules each of $H_2$ and $O_2$. Which one is the limiting reactant? How much excess reactant is left unreacted? What would happen if molecules of $O_2$ were doubled?** **A:** * **Reaction:** $$ 2H_2 + O_2 \rightarrow 2H_2O $$ * **Limiting Reactant:** The ratio is 2 molecules of $H_2$ to 1 molecule of $O_2$. * 500 molecules of $H_2$ require $500/2 = 250$ molecules of $O_2$. We have 500 molecules of $O_2$, so $O_2$ is in excess. * **$H_2$ is the limiting reactant.** * **Product Formed:** The amount of product is determined by the limiting reactant. The ratio of $H_2$ to $H_2O$ is 2:2 (or 1:1). * Therefore, **500 molecules of $H_2O$** will be produced. * **Excess Reactant:** * Initial $O_2$ = 500 molecules. * Reacted $O_2$ = 250 molecules. * Excess $O_2$ = $500 - 250 = \mathbf{250}$ **molecules**. * **If $O_2$ were doubled:** If we had 1000 molecules of $O_2$, $H_2$ would still be the limiting reactant, and the outcome would be exactly the same. 500 molecules of $H_2O$ would be produced. **4. Q: Excess of $AgNO_3$ was added to $FeCl_3$ and 2.02 g of AgCl was produced. Calculate amount (mass) of $FeCl_3$ in the reaction.** **A:** 1. **Reaction:** $$ 3AgNO_3(aq) + FeCl_3(aq) \rightarrow 3AgCl(s) + Fe(NO_3)_3(aq) $$ 2. **Calculate moles of AgCl produced:** * Molar Mass of AgCl = $107.87 + 35.45 = 143.32 \, g/mol$ $$ \text{moles of AgCl} = \frac{2.02 \, g}{143.32 \, g/mol} \approx 0.0141 \, \text{mol} $$ 3. **Use stoichiometry to find moles of $FeCl_3$:** The molar ratio of $FeCl_3$ to $AgCl$ is 1:3. $$ \text{moles of } FeCl_3 = 0.0141 \, \text{mol AgCl} \times \left( \frac{1 \, \text{mol } FeCl_3}{3 \, \text{mol AgCl}} \right) \approx 0.0047 \, \text{mol} $$ 4. **Calculate mass of $FeCl_3$:** * Molar Mass of $FeCl_3 = 55.85 + 3(35.45) = 162.2 \, g/mol$ $$ \text{mass of } FeCl_3 = 0.0047 \, \text{mol} \times 162.2 \, g/mol \approx \mathbf{0.762 \, g} $$

4.2 Limiting And Non Limiting Reactants | Stoichiometry | Chemistry 11