Question

Methane \(\left(\mathrm{CH}_{4}\right)\) is burned in the presence of diatomic oxygen. The combustion products consist of water vapor and carbon dioxide gas. Determine the mass of water vapor generated when 1 lbm of methane is burned.

Short answer

The mass of water vapor generated when 1 lbm of methane is burned is 2.25 lbm.

Step-by-step solution

Write down the balanced chemical equation

The combustion reaction of methane with diatomic oxygen can be represented by the following balanced chemical equation: $$\mathrm{CH}_{4} + 2\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + 2\mathrm{H}_{2}\mathrm{O}$$

Convert the given mass to moles

We are given that 1 lbm (1 pound-mass) of methane is burned. To work with this quantity, we need to convert it from mass to moles. To do this, we'll use the molar mass of methane, which is 16 g/mol (12 g/mol for carbon and 4 g/mol for hydrogen). First, we need to convert 1 lbm to grams. There are 453.592 grams in 1 lbm (1 pound-mass), so we have: $$1 \mathrm{lbm} = 453.592 \mathrm{g}$$ Now, we can convert grams to moles using the molar mass of methane (\(\mathrm{CH}_{4}\)): $$\text{moles of}\: \mathrm{CH}_{4} = \frac{453.592 \mathrm{g}}{16 \mathrm{g/mol}} = 28.35 \mathrm{mol}$$

Apply stoichiometry to find moles of water vapor produced

From the balanced chemical equation, we can see that the stoichiometric ratio between methane and water vapor is 1:2, which means that for every mole of methane burned, 2 moles of water vapor are formed. So, from the moles of methane burned, we can calculate the moles of water vapor formed: $$\text{moles of}\: \mathrm{H}_{2}\mathrm{O} = 28.35 \mathrm{mol\:CH}_{4} \times \frac{2\mathrm{mol\: H}_{2}\mathrm{O}}{1\mathrm{mol\: CH}_{4}} = 56.7 \mathrm{mol\: H}_{2}\mathrm{O}$$

Convert moles of water vapor to mass

Now that we know the moles of water vapor produced, we can convert it back to mass using the molar mass of water, which is 18 g/mol (16 g/mol for oxygen and 2 g/mol for hydrogen): $$\text{mass of}\: \mathrm{H}_{2}\mathrm{O} = 56.7 \mathrm{mol\: H}_{2}\mathrm{O} \times 18 \mathrm{g/mol} = 1020.6 \mathrm{g}$$

Convert mass of water vapor to pound-mass

Finally, we need to convert the mass of water vapor from grams back to pound-mass (lbm) to obtain the answer: $$\text{mass of}\: \mathrm{H}_{2}\mathrm{O} = \frac{1020.6 \mathrm{g}}{453.592 \mathrm{g/lbm}} = 2.25 \mathrm{lbm}$$

Final Answer

The mass of water vapor generated when 1 lbm of methane is burned is 2.25 lbm.

Key concepts

Stoichiometry

Understanding stoichiometry is crucial when examining chemical reactions, such as the combustion of methane. Stoichiometry is the part of chemistry that deals with the relative quantities of reactants and products in chemical reactions.

For the combustion of methane, stoichiometry allows us to predict the amount of product formed from a given quantity of reactant, or methane in this case. The balanced chemical equation for this combustion process is a direct representation of stoichiometry in action: for every one mole of methane (\(CH_{4}\)), two moles of water (\(H_{2}O\)) are produced.

To carry out stoichiometric calculations, we establish a mole ratio from the balanced equation, which in the example is 1:2 for methane to water. Applying this proportion, we can determine the mass of water formed from a known mass of methane by first converting the mass of methane to moles. Therefore, if we know the amount of one substance in a reaction, we can calculate the amount of any other substance in the reaction using the stoichiometric ratios.

Molar Mass

When calculating the amount of substances involved in a chemical reaction, the concept of molar mass is essential. Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol).

For our methane combustion example, knowing the molar masses of methane and water is pivotal. Methane has a molar mass of 16 g/mol, which consists of the sum of the atomic masses of carbon (12 g/mol) and hydrogen (4 g/mol for four atoms). Similarly, the molar mass of water (\(H_{2}O\)) is 18 g/mol, with oxygen contributing 16 g/mol and hydrogen contributing 2 g/mol for the two atoms.

The molar mass enables the conversion from mass to moles and vice versa. When given the mass of methane burned, we use its molar mass to find the corresponding number of moles, which in turn lets us calculate how much water vapor, in moles, is produced.

Chemical Reaction Balancing

Balancing chemical equations is a key skill to ensure the law of conservation of mass is adhered to during a chemical reaction. A balanced equation has equal numbers of each atom on both sides of the reaction.

In the case of methane combustion, the balanced chemical equation is \(CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O\). Here, we see that there are four hydrogen atoms and one carbon atom from methane and two oxygen molecules from diatomic oxygen resulting in one carbon dioxide molecule and two water molecules. Without balancing, it would be impossible to correctly apply stoichiometry as the mole ratios of the reactants and products would be inaccurate.

This process of balancing forms the cornerstone for all subsequent calculations, including stoichiometry and mass-mole conversions. A properly balanced equation reflects the actual chemical process and allows accurate prediction of product formation.

Mass-Mole Conversion

Mass-mole conversion is an essential step in solving many chemistry problems, including those involving the combustion of methane.

The process involves using the substance's molar mass to convert between the mass of a substance in grams and the amount of substance in moles. In the methane combustion example, we convert the given mass of methane (in pounds) to grams, and then use methane’s molar mass to convert that mass to moles. Later, stoichiometry helps us determine the moles of water produced, which we then convert to grams using water’s molar mass, and finally back to pounds for the final answer.

In essence, the mass-mole conversion is a two-step process where molar mass serves as the conversion factor, bridging the gap between the macroscopic scale (mass) and the microscopic scale (moles) of a substance.