Question

To produce 15.4 L of nitrogen dioxide at 310 K and 2.0 atm, how many liters of nitrogen gas and oxygen gas are required?

Short answer

To produce 15.4 L of nitrogen dioxide at 310 K and 2.0 atm, we need 7.68 L of nitrogen gas and 15.4 L of oxygen gas.

Step-by-step solution

Write the balanced chemical equation.

For the reaction between nitrogen gas (N₂) and oxygen gas (O₂) to produce nitrogen dioxide (NO₂), the balanced chemical equation is: \[ N₂(g) + 2O₂(g) \rightarrow 2NO₂(g) \]

Calculate the number of moles of nitrogen dioxide.

Using the ideal gas law, we can find the number of moles of nitrogen dioxide. The ideal gas law is given by: \[ PV = nRT \] Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. From the exercise, the volume of NO₂ is 15.4 L, the pressure is 2.0 atm, and the temperature is 310 K. We want to calculate the number of moles (n) of NO₂ produced, given that R = 0.0821 L·atm/mol·K. Rearrange and solve for n: \[ n = \frac{PV}{RT} \] Plug in given values: \( n = \frac{(2.0\,atm)(15.4\,L)}{(0.0821\,L\,atm/mol\,K)(310\,K)} \) Calculate: \( n = 1.210\,moles\,NO₂ \)

Use stoichiometry to find the number of moles of N₂ and O₂ required.

From the balanced chemical equation, we know the following mole ratios: - 1 mole of N₂ reacts with 2 moles of O₂ to produce 2 moles of NO₂. We have calculated 1.210 moles of NO₂. Now, we can use stoichiometry to find the number of moles of N₂ and O₂ required based on the mole ratios: For N₂: \( \frac{1\,moles\,N₂}{2\,moles\,NO₂} = \frac{x\,moles\,N₂}{1.210\,moles\,NO₂} \) For O₂: \( \frac{2\,moles\,O₂}{2\,moles\,NO₂} = \frac{y\,moles\,O₂}{1.210\,moles\,NO₂} \) Solve for x and y to find the moles of N₂ and O₂ required: \( x = 0.605\,moles\,N₂ \) \( y = 1.210\,moles\,O₂ \)

Calculate the volume of N₂ and O₂ required.

Now that we have the number of moles of N₂ and O₂ required, we can use the ideal gas law again to find the volume of each gas at the given temperature and pressure conditions. For N₂: \( V = \frac{nRT}{P} = \frac{(0.605\,moles)(0.0821\,L\,atm/mol\,K)(310\,K)}{2.0\,atm} = 7.68\,L\,N₂ \) For O₂: \( V = \frac{nRT}{P} = \frac{(1.210\,moles)(0.0821\,L\,atm/mol\,K)(310\,K)}{2.0\,atm} = 15.4\,L\,O₂ \) Hence, to produce 15.4 L of nitrogen dioxide at the given conditions, we need 7.68 L of nitrogen gas and 15.4 L of oxygen gas.

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that allows us to calculate the quantities of reactants or products involved in a chemical reaction. This involves using ratios from the balanced chemical equation to relate different substances.
This concept is crucial for answering questions like the one in our exercise, where the task is to determine how much of each reactant is needed to produce a specific amount of product.
In our specific case, we determine the moles required for nitrogen (\(N_2\))and oxygen (\(O_2\))to produce nitrogen dioxide (\(NO_2\))using the stoichiometric relationships derived from the balanced chemical equation:

• 1 mole of nitrogen gas reacts with 2 moles of oxygen gas to form 2 moles of nitrogen dioxide.

By knowing how many moles of (\(NO_2\))are produced and using stoichiometry, you can easily find out how many moles of (\(N_2\))and (\(O_2\))are needed.

Chemical Reactions

A chemical reaction involves the transformation of reactants into products. Each chemical reaction is distinct and follows a specific pathway. In our exercise, nitrogen gas (\(N_2\))and oxygen gas (\(O_2\))combine to form nitrogen dioxide (\(NO_2\)). This is an example of a synthesis reaction where simpler substances combine to form a more complex product.
Reactions occur when molecules collide with sufficient energy and in the correct orientation, leading to a change in their chemical bonds. The energy required for a reaction to take place is known as the activation energy.
Understanding how chemical reactions work is crucial for predicting the outcome of a reaction, calculating the amounts of products formed, or the reactants required.

Gas Laws

Gas laws describe the behavior of gases in different conditions of pressure, volume, and temperature. In this exercise, we apply the Ideal Gas Law, which is fundamental in understanding how gases behave.
The Ideal Gas Law is expressed as:\( PV = nRT \)where:

• \(P\) is pressure,
• \(V\) is volume,
• \(n\) is the number of moles of gas,
• \(R\) is the ideal gas constant, and
• \(T\) is temperature in Kelvin.

This law helps us calculate one variable if the others are known. Here, it is used to find the number of moles of nitrogen dioxide produced and then the volumes of the reactants required. The assumptions of the ideal gas law apply best under conditions of low pressure and high temperature.

Balanced Chemical Equations

A balanced chemical equation accurately represents a chemical reaction and shows the relationship between reactants and products. In this exercise, balancing the equation ensures that the law of conservation of mass is obeyed, meaning matter is neither created nor destroyed during the reaction.
The balanced equation:\[ N_2(g) + 2O_2(g) \rightarrow 2NO_2(g) \]illustrates that:

• One molecule of nitrogen reacts with two molecules of oxygen to produce two molecules of nitrogen dioxide.

Balancing an equation is a critical skill in chemistry, as it sets the framework for stoichiometry calculations.
To balance an equation, the number of atoms for each element on the reactants side must match the number of atoms on the products side. This process often involves adjusting coefficients to achieve this balance.