Question

When nitrogen gas \(\left(\mathrm{N}_{2}\right)\) is reacted with \(\mathrm{H}_{2}\) under certain conditions, the following reaction occurs: $$\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightarrow 2 \mathrm{NH}_{3}$$ Based on the results of Experiment \(1,\) if 10 \(\mathrm{mL}\) of \(\mathrm{N}_{2}\) were completely reacted with 40 \(\mathrm{mL}\) of \(\mathrm{H}_{2}\) at the same pressure and temperature, what volume of \(\mathrm{H}_{2}\) would remain unreacted? F. \(0 \mathrm{mL}\) G. 10 \(\mathrm{mL}\) H. 20 \(\mathrm{mL}\) J. 30 \(\mathrm{mL}\)

Short answer

Answer: 10 mL

Step-by-step solution

1. Write the balanced chemical equation

The given balanced chemical equation is: N₂ + 3H₂ → 2NH₃

2. Identify the given volumes of gases

From the problem, we are given the following volumes of gases: Volume of N₂ (V(N₂)) = 10 mL Volume of H₂ (V(H₂)) = 40 mL

3. Determine the limiting reactant

For every mole of N₂, we need 3 moles of H₂ to complete the reaction. We can find the ratio of moles of H₂ required for the given moles (volume) of N₂: Ratio = Volume of H₂ required / Volume of N₂ Ratio = 3 Now, let's calculate the volume of H₂ required for the given volume of N₂: Volume of H₂ required = Ratio × Volume of N₂ = 3 × 10 mL = 30 mL The actual available volume of H₂ is 40 mL, which is more than the required volume of 30 mL. Thus, N₂ is the limiting reactant, and some H₂ will remain unreacted.

4. Calculate the volume of unreacted H₂

Since we know the volume of H₂ required for the reaction and the available volume of H₂, we can calculate the volume of unreacted H₂: Volume of unreacted H₂ = Available Volume of H₂ - Volume of H₂ required Volume of unreacted H₂ = 40 mL - 30 mL = 10 mL Therefore, the volume of unreacted H₂ is 10 mL, which corresponds to answer choice G.

Key concepts

Balanced Chemical Equations

Understanding balanced chemical equations is crucial in chemistry. It's similar to making sure you have enough ingredients to bake a cake; if the recipe isn't balanced, your cake won't turn out right. In chemical terms, a balanced equation reflects the law of conservation of mass, stating that matter is neither created nor destroyed in a chemical reaction. Therefore, you must have the same number of each type of atom on both sides of the reaction.

In the given problem, the balanced equation is: \[ \mathrm{N}_{2} + 3 \mathrm{H}_{2} \rightarrow 2 \mathrm{NH}_{3} \]. This indicates that one molecule of nitrogen gas (\( \mathrm{N}_{2} \)) reacts with three molecules of hydrogen gas (\( \mathrm{H}_{2} \)) to produce two molecules of ammonia (\( \mathrm{NH}_{3} \)). The coefficients in front of each compound—1 for \( \mathrm{N}_{2} \), 3 for \( \mathrm{H}_{2} \), and 2 for \( \mathrm{NH}_{3} \)—ensure the equation is balanced, with an equal number of nitrogen and hydrogen atoms on both sides.

Limiting Reactant

The concept of a limiting reactant is akin to running out of flour while baking; it doesn't matter how much of the other ingredients you have left, without enough flour, you can't make more cake. In chemical reactions, the limiting reactant is the substance that is totally consumed when the chemical reaction is complete, thus determining the maximum amount of product that can be formed.

The exercise demonstrates this concept by comparing the volumes of \( \mathrm{N}_{2} \) and \( \mathrm{H}_{2} \) provided. Since \( \mathrm{N}_{2} \) requires thrice the volume of \( \mathrm{H}_{2} \) for a complete reaction, we find that \( \mathrm{N}_{2} \) is the limiting reactant. Therefore, no matter how much \( \mathrm{H}_{2} \) you start with, the reaction will stop when all of the \( \mathrm{N}_{2} \) has been used up, indicating that \( \mathrm{N}_{2} \) controls the extent of the reaction.

Mole Ratio

The mole ratio serves as a bridge between reactants in a balanced chemical equation much like a recipe specifies the ratio of cups of flour to eggs. It tells us the proportion in which reactants combine and products form. The mole ratio is derived directly from the coefficients in a balanced equation and is essential for calculations involving relative quantities of reactants and products.

In our exercise, the mole ratio is the anchor that helps us determine how much \( \mathrm{H}_{2} \) will be left over. As per the balanced equation, the mole ratio of \( \mathrm{H}_{2} \) to \( \mathrm{N}_{2} \) is 3:1. This means for every 10 mL of \( \mathrm{N}_{2} \), we require 30 mL of \( \mathrm{H}_{2} \) (3 times the volume of \( \mathrm{N}_{2} \)). Knowing that, we can calculate the leftover volume of \( \mathrm{H}_{2} \) by subtracting the volume required by the reaction from the initial volume provided.