Problem 11 - Chemistry: An Atoms-Focused Approach

Question

Depending on reaction conditions, \(\mathrm{O}_{2}\) may combine with \(\mathrm{N}_{2}\) to form \(\mathrm{NO}\) or \(\mathrm{NO}_{2} .\) If \(x\) grams of \(\mathrm{O}_{2}\) combine with \(y\) grams of \(\mathrm{N}_{2}\) to form \(\mathrm{NO}\), how many grams of \(\mathrm{O}_{2}\) combine with \(y\) grams of \(\mathrm{N}_{2}\) to form \(\mathrm{NO}_{2} ?\)

Step-by-Step Solution

Answer

Answer: 16 grams.

Step 1: Identify the balanced chemical equations

The chemical reactions are: (i) \(\mathrm{O}_{2} + \mathrm{N}_{2} \rightarrow 2\mathrm{NO}\) (ii) \(\mathrm{O}_{2} + 2\mathrm{N}_{2} \rightarrow 2\mathrm{NO}_{2}\)

Step 2: Compute the moles of each element

We are given \(x\) grams of \(\mathrm{O}_{2}\) and \(y\) grams of \(\mathrm{N}_{2}\) as the starting materials. We need to compute the moles of each given their molar mass: The molar mass of \(\mathrm{O}_{2}\) = \(2 \times 16 = 32\) g/mol The molar mass of \(\mathrm{N}_{2}\) = \(2 \times 14 = 28\) g/mol

Step 3: Find the mole ratio between O₂ and N₂ in each reaction

For the equation (i), the mole ratio between \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) is 1:1. For the equation (ii), the mole ratio between \(\mathrm{O}_{2}\) and \(\mathrm{N}_{2}\) is 1:2.

Step 4: Calculate the moles of O₂ needed to form NO₂

From the equation (i): Moles of \(\mathrm{O}_{2}\) = \(\frac{x}{32}\) and moles of \(\mathrm{N}_{2}\) = \(\frac{y}{28}\) Since the mole ratio is 1:1, we have: $$\frac{x}{32} = \frac{y}{28} $$ From the equation (ii): Let the amount of \(\mathrm{O}_{2}\) needed to combine with \(y\) grams of \(\mathrm{N}_{2}\) to form \(\mathrm{NO}_{2}\) be \(z\) grams. Moles of \(\mathrm{O}_{2}\) = \(\frac{z}{32}\) and moles of \(\mathrm{N}_{2}\) = \(\frac{y}{28}\) Since the mole ratio is 1:2, we have: $$\frac{z}{32} = \frac{1}{2} \times \frac{y}{28} $$

Step 5: Solve for z

Now we have two equations with two variables - \(x\) and \(z\). $$\frac{x}{32} = \frac{y}{28} $$ and $$\frac{z}{32} = \frac{1}{2} \times \frac{y}{28} $$ From the first equation, we can find the value of \(x\) in terms of \(y\): $$x = 32\times\frac{y}{28}$$ Now, substitute this value of \(x\) in the second equation: $$\frac{z}{32} = \frac{1}{2} \times \frac{32\times\frac{y}{28}}{28}$$ After simplifying the equation, we get: $$z = \frac{32}{2} = 16$$ Thus, 16 grams of \(\mathrm{O}_{2}\) combine with \(y\) grams of \(\mathrm{N}_{2}\) to form \(\mathrm{NO}_{2}\).

Key Concepts

Mole Ratio

The concept of mole ratio is crucial in understanding chemical reactions. It tells us the proportion of moles among the reactants and products that participate in a reaction. In other words, it shows how much of one substance reacts with another to form products.

For the chemical reactions given in the exercise:

• In the first reaction, \( \mathrm{O}_{2} + \mathrm{N}_{2} \rightarrow 2\mathrm{NO} \), the mole ratio of \(\mathrm{O}_{2}\) to \(\mathrm{N}_{2}\) is 1:1, meaning one mole of \(\mathrm{O}_{2}\) reacts with one mole of \(\mathrm{N}_{2}\) to produce two moles of \(\mathrm{NO}\).
• In the second reaction, \( \mathrm{O}_{2} + 2\mathrm{N}_{2} \rightarrow 2\mathrm{NO}_{2} \), the mole ratio is 1:2. This means that one mole of \(\mathrm{O}_{2}\) must react with two moles of \(\mathrm{N}_{2}\) to form two moles of \(\mathrm{NO}_{2}\).

Understanding mole ratio helps in predicting the amount of substances consumed or produced in a reaction. It's derived from the coefficients of the balanced chemical equations, making it an integral part of reaction stoichiometry.

Balanced Chemical Equations

Balanced chemical equations are vital to chemists as they accurately depict the conservation of mass during a reaction.

The law of conservation of mass states that atoms cannot be created or destroyed. Therefore, a balanced equation must reflect the same number and type of atoms on both sides of the equation. For the reactions in the exercise, the equations are:

• \( \mathrm{O}_{2} + \mathrm{N}_{2} \rightarrow 2\mathrm{NO} \)
• \( \mathrm{O}_{2} + 2\mathrm{N}_{2} \rightarrow 2\mathrm{NO}_{2} \)

To balance an equation, you adjust the coefficients (the numbers in front of the molecules) to ensure equal numbers of each type of atom are present on both sides. Each atom's count must match since no atoms are lost or gained during the reaction.
When equations are balanced, they can be used to calculate amounts of reactants needed or products formed using mole ratios. This accurate prediction is the essence of chemical reaction stoichiometry.

Molar Mass

Molar mass is the mass of one mole of a given substance. It's expressed in grams per mole (g/mol), and it’s essential for converting between the mass of a substance and the number of moles.

To find the molar mass of a molecule, sum the atomic masses of all the atoms in the molecule. Here's how it's done with the molecules from the exercise:

• \( \mathrm{O}_{2} \): Each oxygen atom has an atomic mass of 16 g/mol. Therefore, the molar mass of \( \mathrm{O}_{2} \) is \( 2 \times 16 = 32 \) g/mol.
• \( \mathrm{N}_{2} \): Each nitrogen atom has an atomic mass of 14 g/mol. Thus, the molar mass of \( \mathrm{N}_{2} \) is \( 2 \times 14 = 28 \) g/mol.

Understanding molar mass allows us to convert masses of substances to the number of moles, using the formula: \[\text{Number of moles} = \frac{\text{Mass of Substance}}{\text{Molar Mass}}\]This relationship is crucial for calculating how much reactant you need or how much product you'll get in a chemical reaction. It's also fundamental for using balanced chemical equations to determine mole ratios accurately.