Number of Moles of Water Produced

Oct 13, 2025 by sorusgentrification.com 3 views

Introduction

The combustion of butane ( $C_4H_{10}$ ) is a highly exothermic reaction that produces carbon dioxide ( $CO_2$ ) and water ( $H_2O$ ) as products. In this article, we will calculate the volume of water produced by the combustion of 10.0 grams of butane at 35°C and 0.95 atm.

Chemical Equation

The balanced chemical equation for the combustion of butane is:

$C_4H_{10} + 6.5O_2 \rightarrow 4CO_2 + 5H_2O$

Molar Mass of Butane

The molar mass of butane ( $C_4H_{10}$ ) is calculated as follows:

Molar mass of $C_4H_{10}$ = (4 x 12.01 g/mol) + (10 x 1.008 g/mol) = 48.04 g/mol + 10.08 g/mol = 58.12 g/mol

Number of Moles of Butane

The number of moles of butane can be calculated using the formula:

Number of moles = mass of butane / molar mass of butane = 10.0 g / 58.12 g/mol = 0.172 mol

Limiting Reactant

To determine the limiting reactant, we need to calculate the number of moles of oxygen required for the combustion of butane.

Number of moles of oxygen required = (number of moles of butane x 6.5) = 0.172 mol x 6.5 = 1.12 mol

Number of Moles of Water Produced

The number of moles of water produced can be calculated using the formula:

Number of moles of water produced = (number of moles of butane x 5) = 0.172 mol x 5 = 0.86 mol

Volume of Water Produced

To calculate the volume of water produced, we need to use the ideal gas law:

PV = nRT

where: P = pressure (0.95 atm) V = volume (in liters) n = number of moles (0.86 mol) R = gas constant (0.0821 L atm/mol K) T = temperature (35°C + 273 = 308 K)

Rearranging the equation to solve for V, we get:

V = nRT / P = (0.86 mol x 0.0821 L atm/mol K x 308 K) / 0.95 atm = 2.33 L

Conclusion

In this article, we calculated the volume of water produced by the combustion of 10.0 grams of butane at 35°C and 0.95 atm. The results show that the volume of water produced is 2.33 liters.

References

[1] CRC Handbook of Chemistry and Physics, 97th Edition
[2] Atkins, P. W., & de Paula, J. (2010). Physical Chemistry. Oxford University Press.

Additional Information

The combustion of butane is a highly exothermic reaction that releases a large amount of energy.
The reaction is also highly efficient, with a high yield of water and carbon dioxide.
The ideal gas law can be used to calculate the volume of gases produced in a reaction.

Calculations

Molar mass of butane: 58.12 g/mol
Number of moles of butane: 0.172 mol
Number of moles of oxygen required: 1.12 mol
Number of moles of water produced: 0.86 mol
Volume of water produced: 2.33 L

Questions and Answers

Q: What is the molar mass of butane? A: The molar mass of butane is 58.12 g/mol.
Q: How many moles of butane are required for the combustion reaction? A: 0.172 mol of butane are required for the combustion reaction.
Q: How many moles of water are produced in the combustion reaction? A: 0.86 mol of water are produced in the combustion reaction.
Q: What is the volume of water produced in the combustion reaction? A: The volume of water produced is 2.33 L.

Introduction

In our previous article, we calculated the volume of water produced by the combustion of 10.0 grams of butane at 35°C and 0.95 atm. In this article, we will answer some frequently asked questions related to the combustion of butane and the calculation of the volume of water produced.

Q: What is the molar mass of butane?

A: The molar mass of butane ( $C_4H_{10}$ ) is 58.12 g/mol.

Q: How many moles of butane are required for the combustion reaction?

A: 0.172 mol of butane are required for the combustion reaction.

Q: How many moles of water are produced in the combustion reaction?

A: 0.86 mol of water are produced in the combustion reaction.

Q: What is the volume of water produced in the combustion reaction?

A: The volume of water produced is 2.33 L.

Q: What is the ideal gas law and how is it used in this calculation?

A: The ideal gas law is a mathematical equation that describes the behavior of ideal gases. It is used to calculate the volume of gases produced in a reaction. In this calculation, we used the ideal gas law to calculate the volume of water produced.

Q: What is the significance of the temperature and pressure in this calculation?

A: The temperature and pressure are important factors in this calculation because they affect the volume of the gas produced. In this calculation, we used a temperature of 35°C and a pressure of 0.95 atm.

Q: What is the limiting reactant in this reaction?

A: The limiting reactant in this reaction is butane. This means that butane is the reactant that is consumed first in the reaction.

Q: What is the balanced chemical equation for the combustion of butane?

A: The balanced chemical equation for the combustion of butane is:

$C_4H_{10} + 6.5O_2 \rightarrow 4CO_2 + 5H_2O$

Q: What is the molar mass of oxygen?

A: The molar mass of oxygen is 32.00 g/mol.

Q: How many moles of oxygen are required for the combustion reaction?

A: 1.12 mol of oxygen are required for the combustion reaction.

Q: What is the significance of the gas constant (R) in this calculation?

A: The gas constant (R) is a mathematical constant that is used in the ideal gas law. It is used to calculate the volume of gases produced in a reaction.

Q: What is the significance of the number of moles (n) in this calculation?

A: The number of moles (n) is the amount of substance that is used in the reaction. In this calculation, we used the number of moles of butane and oxygen to calculate the volume of water produced.

Q: What is the significance of the temperature (T) in this calculation?

A: The temperature (T) is an important factor in this calculation because it affects the volume of the gas produced. In this calculation, we used a temperature of 35°C.

Q: What is the significance of the pressure (P) in this calculation?

A: The pressure (P) is an important factor in this calculation because it affects the volume of the gas produced. In this calculation, we used a pressure of 0.95 atm.

Q: What is the significance of the volume (V) in this calculation?

A: The volume (V) is the amount of space that is occupied by the gas produced in the reaction. In this calculation, we used the volume of water produced to answer the question.