Problem 51 - Chemistry The Molecular Science

Question

Determine the amount of reaction (in moles) that takes place for each process $$ 2 \mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{3}(\mathrm{~g}) $$ (a) \(2 \mathrm{~mol} \mathrm{O}_{2}\) reacts (b) \(0.115 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{3}\) forms (c) \(4.73 \mathrm{~g}\) NO reacts

Step-by-Step Solution

Answer

(a) 2 mol O2 yields 8 mol NO and 4 mol N2O3; (b) 0.115 mol N2O3 needs 0.23 mol NO; (c) 4.73 g NO is 0.1577 mol, producing 0.0394 mol O2 and 0.0789 mol N2O3.

Step 1: Understanding the Chemical Equation

The given chemical reaction is \( 2 \mathrm{NO} (\mathrm{g}) + \frac{1}{2} \mathrm{O}_2 (\mathrm{g}) \rightarrow \mathrm{N}_2 \mathrm{O}_3 (\mathrm{g}) \). This means every 2 moles of \( \mathrm{NO} \) react with \( \frac{1}{2} \) mole of \( \mathrm{O}_2 \) to produce 1 mole of \( \mathrm{N}_2 \mathrm{O}_3 \). The stoichiometric coefficients show the mole-to-mole relationships between reactants and products.

Step 2: Calculating for O2 in Part (a)

Given: 2 moles of \( \mathrm{O}_2 \). According to the reaction, 1 mole of \( \mathrm{O}_2 \) corresponds to 4 moles of \( \mathrm{NO} \) and produces 2 moles of \( \mathrm{N}_2 \mathrm{O}_3 \). Therefore, 2 moles of \( \mathrm{O}_2 \) require and produce twice as much: \[ 2 \text{ moles } \mathrm{O}_2 \times \frac{4 \text{ moles } \mathrm{NO}}{1 \text{ mole } \mathrm{O}_2} = 8 \text{ moles } \mathrm{NO} \] \[ 2 \text{ moles } \mathrm{O}_2 \times \frac{2 \text{ moles } \mathrm{N}_2\mathrm{O}_3}{1 \text{ mole } \mathrm{O}_2} = 4 \text{ moles } \mathrm{N}_2\mathrm{O}_3 \]

Step 3: Calculating for N2O3 in Part (b)

Given: 0.115 moles of \( \mathrm{N}_2 \mathrm{O}_3 \) formed. From the reaction, 1 mole of \( \mathrm{N}_2 \mathrm{O}_3 \) comes from 2 moles of \( \mathrm{NO} \), so: \[ 0.115 \text{ moles } \mathrm{N}_2\mathrm{O}_3 \times \frac{2 \text{ moles } \mathrm{NO}}{1 \text{ mole } \mathrm{N}_2\mathrm{O}_3} = 0.23 \text{ moles } \mathrm{NO} \] Also, for \( \mathrm{O}_2 \): \[ 0.115 \text{ moles } \mathrm{N}_2\mathrm{O}_3 \times \frac{0.5 \text{ moles } \mathrm{O}_2}{1 \text{ mole } \mathrm{N}_2\mathrm{O}_3} = 0.0575 \text{ moles } \mathrm{O}_2 \]

Step 4: Converting Mass to Moles in Part (c)

Given: 4.73 g of \( \mathrm{NO} \). First, calculate the moles of \( \mathrm{NO} \): The molar mass of \( \mathrm{NO} \) is \( 14 \text{ (N) } + 16 \text{ (O) } = 30 \text{ g/mol} \). \[ \frac{4.73 \text{ g }}{30 \text{ g/mol }} = 0.1577 \text{ moles of } \mathrm{NO} \] From the stoichiometry of the equation, \( \frac{1}{2} \) mole of \( \mathrm{O}_2 \) reacts with 2 moles of \( \mathrm{NO} \), so for \( 0.1577 \) moles of \( \mathrm{NO} \): \[ 0.1577 \text{ moles } \mathrm{NO} \times \frac{1 \text{ mole } \mathrm{O}_2}{4 \text{ moles } \mathrm{NO}} = 0.0394 \text{ moles } \mathrm{O}_2 \] And it forms: \[ 0.1577 \text{ moles } \mathrm{NO} \times \frac{1 \text{ mole } \mathrm{N}_2\mathrm{O}_3}{2 \text{ moles } \mathrm{NO}} = 0.0789 \text{ moles } \mathrm{N}_2\mathrm{O}_3 \]

Key Concepts

Chemical Reactions

Chemical reactions are the processes where substances are transformed into different substances. This transformation happens when reactants interact to form products. Every chemical reaction follows the law of conservation of mass. This means the mass of the reactants will always equal the mass of the products.

In our exercise, the reaction involves nitrogen monoxide (NO) and oxygen ( O_2 ) as reactants. They are converted into dinitrogen trioxide ( N_2O_3 ) as the product. Reactions are represented by chemical equations which show the substances involved.

• The reactants are on the left side.
• The product is on the right side.
• An arrow points from reactants to products, indicating the direction of the reaction.

Chemical reactions are fundamental to stoichiometry. They allow us to understand how substances interconvert. Stoichiometry is the math that helps us calculate the precise amounts needed to fully react materials. It is essential for predicting the yield of a reaction.

Moles Calculation

Moles are a unit of measurement used to count particles at the atomic and molecular level. A mole is defined as exactly 6.022 x 10²³ entities, and this number is known as Avogadro's number. Moles allow chemists to work with the macroscopic amounts of material, for reactions and calculations.

Our exercise requires computing the moles of each substance involved in the reaction, based on the given data. For example, in part (a), we know there are 2 moles of O_2 reacting. With the balanced equation, we calculate how many moles of NO and N_2O_3 will be involved. Similarly, in part (b), based on the moles of N_2O_3 formed, the moles of NO and O_2 consumed are calculated.

• This helps ensure that the reaction components are used optimally.
• Such calculations guide chemical processes in industry and research by predicting amounts of materials.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They convey essential information about the reaction substances, including quantities and states of matter.

Equations must be balanced, meaning they have the same number of each type of atom on both sides. In our example, the equation is:\[2 \text{NO} (\text{g}) + \frac{1}{2} \text{O}_2 (\text{g}) \rightarrow \text{N}_2 \text{O}_3 (\text{g})\]

• The coefficients (the numbers in front of species) indicate the number of moles of each substance.
• This balanced equation tells us:
  • 2 moles of NO are needed for every 0.5 mole of O_2.
  • The reaction produces 1 mole of N_2O_3.
• The subscripts in the molecule denote the number of atoms of each element in the molecule. For instance, in N_2O_3, there are two nitrogen (N) atoms and three oxygen (O) atoms.

Understanding how to read and interpret chemical equations is crucial for solving stoichiometry problems. It helps predict the outcomes of reactions in practical applications.