Question

Suppose that reactions (a) and (b) each have a \(92 \%\) yield. Starting with \(112 \mathrm{g} \mathrm{CH}_{4}\) in reaction \((\mathrm{a})\) and an excess of \(\mathrm{Cl}_{2}(\mathrm{g}),\) how many grams of \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) are formed in reaction (b)? (a) \(\mathrm{CH}_{4}+\mathrm{Cl}_{2} \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}+\mathrm{HCl}\) (b) \(\mathrm{CH}_{3} \mathrm{Cl}+\mathrm{Cl}_{2} \longrightarrow \mathrm{CH}_{2} \mathrm{Cl}_{2}+\mathrm{HCl}\)

Short answer

The mass of \(\mathrm{CH}_{2}\mathrm{Cl}_{2}\) formed in reaction (b) is estimated to be around \(501.67\) g.

Step-by-step solution

Calculate the number of moles of CH4

The initial mass of \(\mathrm{CH}_{4}\) is given as 112 g. The molar mass of \(\mathrm{CH}_{4}\) is approximately 16.04 g/mol. Therefore, the initial amount of \(\mathrm{CH}_{4}\) in moles is \(moles = \frac{mass}{molar\,mass}\) = \(\frac{112\, g}{16.04 \,g/mol} = 6.98 \,moles\)

Calculate the number of moles of CH3Cl produced in reaction (a)

The yield of reaction (a) is 92%, thus the actual amount of \(\mathrm{CH}_{3}\mathrm{Cl}\) produced is \((6.98 moles)*0.92 = 6.42 moles\)

Calculate the number of moles of CH2Cl2 produced in reaction (b)

The yield of reaction (b) is also 92%, thus the actual amount of CH2Cl2 that is produced is \((6.42 moles)*0.92 = 5.91 moles\)

Convert the number of moles of CH2Cl2 to mass

The molar mass of \(\mathrm{CH}_{2}\mathrm{Cl}_{2}\) is approximately 84.93 g/mol. Therefore, the final amount of \(\mathrm{CH}_{2}\mathrm{Cl}_{2}\) in grams is \(grams = moles * molar\,mass = (5.91\, moles) * (84.93 \,g/mol) = 501.67 g\)

Key concepts

Understanding Chemical Reactions

Chemical reactions involve the transformation of reactants into products. In our exercise, we focus on two reactions. The first reaction processes methane (\(\mathrm{CH}_4\)) with chlorine (\(\mathrm{Cl}_2\)) to produce chloromethane (\(\mathrm{CH}_3\mathrm{Cl}\)) and hydrochloric acid (\(\mathrm{HCl}\)). The equation:

• \[\mathrm{CH}_4 + \mathrm{Cl}_2 \rightarrow \mathrm{CH}_3\mathrm{Cl} + \mathrm{HCl}\]

illustrates a substitution reaction, where a hydrogen atom in methane is replaced by a chlorine atom. Reaction (b) is also a substitution reaction but involves chloromethane and chlorine to yield dichloromethane (\(\mathrm{CH}_2\mathrm{Cl}_2\)) and more \(\mathrm{HCl}\):

• \[\mathrm{CH}_3\mathrm{Cl} + \mathrm{Cl}_2 \rightarrow \mathrm{CH}_2\mathrm{Cl}_2 + \mathrm{HCl}\]

Note that excess \(\mathrm{Cl}_2\) ensures the completion of these reactions, pushing the chemical equilibrium toward the formation of desired products.

The Mole Concept and Its Application

The mole concept is integral to stoichiometry as it helps in quantifying substances in chemical reactions. The mole (6.022 x 10²³ entities) is a basic unit in chemistry for measuring the amount of substance.

• In the exercise, you start with 112 g of \(\mathrm{CH}_4\), knowing its molar mass is 16.04 g/mol.
• To find the number of moles: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{112\, \text{g}}{16.04\, \text{g/mol}} = 6.98\, \text{moles}\).

This calculation allows you to proceed to the next reaction steps by relating the amount of \(\mathrm{CH}_4\) to the subsequent products through stoichiometric coefficients from the balanced equations. This ratio serves as a conversion factor linking the initial reactants to potential products.

Reaction Yield and Efficiency

Understanding reaction yield is crucial in predicting the efficiency of chemical processes. The theoretical yield represents the maximum amount of product achievable from given reactants. The actual yield is what you obtain from an experiment, often less due to incomplete reactions or side reactions. The percentage yield quantifies how efficient the reaction is by comparing the actual yield to the theoretical yield.

• Formula for calculating percentage yield:\[\text{Percentage Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100\%\]
• In our step-by-step solution, the 92% yield for both reactions means that only 92% of the maximum possible product is formed.

For example, after the first reaction, you multiply the number of moles of \(\mathrm{CH}_3\mathrm{Cl}\) by 0.92 to find the actual moles produced. This same logic applies to the second reaction with \(\mathrm{CH}_2\mathrm{Cl}_2\). Calculating these yields helps in estimating the quantities of products more accurately for practical applications.