Question

Dichloromethane, \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\), the solvent used to decaffeinate coffee beans, is prepared by reaction of \(\mathrm{CH}_{4}\) with \(\mathrm{Cl}_{2}\). (a) Write the balanced equation. (HCl is also formed.) (b) How many grams of \(\mathrm{Cl}_{2}\) are needed to react with \(50.0 \mathrm{~g}\) of \(\mathrm{CH}_{4} ?\) (c) How many grams of dichloromethane are formed from \(50.0 \mathrm{~g}\) of \(\mathrm{CH}_{4}\) if the percent yield for the reaction is \(76 \% ?\)

Short answer

(a) \(\mathrm{CH}_4 + 2\mathrm{Cl}_2 \rightarrow \mathrm{CH}_2\mathrm{Cl}_2 + 2\mathrm{HCl}\); (b) 441.06 g; (c) 201.18 g.

Step-by-step solution

Write the chemical equation

The reaction between methane (\(\mathrm{CH}_4\)) and chlorine (\(\mathrm{Cl}_2\)) to produce dichloromethane (\(\mathrm{CH}_2\mathrm{Cl}_2\)) and hydrogen chloride (\(\mathrm{HCl}\)) can be written as follows:\[\mathrm{CH}_4 + 2\mathrm{Cl}_2 \rightarrow \mathrm{CH}_2\mathrm{Cl}_2 + 2\mathrm{HCl}\]This equation is balanced as it has the same number of each atom on both sides of the equation.

Calculate moles of CH4

First, we need to calculate how many moles of methane (\(\mathrm{CH}_4\)) there are in 50.0 g. Using the molar mass of methane, which is approximately 16.05 g/mol, we find the number of moles:\[\text{Moles of \(\mathrm{CH}_4\)} = \frac{50.0 \text{ g}}{16.05 \text{ g/mol}} \approx 3.115 \text{ moles}\]

Determine moles of Cl2 needed

According to the balanced equation, 1 mole of \(\mathrm{CH}_4\) requires 2 moles of \(\mathrm{Cl}_2\). Thus, 3.115 moles of \(\mathrm{CH}_4\) will require:\[3.115 \text{ moles } \times 2 \frac{\text{moles of } \mathrm{Cl}_2}{\text{mole of } \mathrm{CH}_4} = 6.230 \text{ moles of } \mathrm{Cl}_2\]

Convert moles of Cl2 to grams

Next, we convert moles of \(\mathrm{Cl}_2\) to grams. The molar mass of \(\mathrm{Cl}_2\) is approximately 70.90 g/mol.\[\text{Grams of } \mathrm{Cl}_2 = 6.230 \text{ moles} \times 70.90 \text{ g/mol} \approx 441.06 \text{ g}\]

Calculate theoretical yield of CH2Cl2

In ideal conditions (100% yield), the moles of \(\mathrm{CH}_2\mathrm{Cl}_2\) can be determined by the same amount of moles of \(\mathrm{CH}_4\) reacted, which is 3.115 moles. The molar mass of \(\mathrm{CH}_2\mathrm{Cl}_2\) is approximately 84.93 g/mol. The theoretical mass is:\[3.115 \text{ moles} \times 84.93 \text{ g/mol} \approx 264.71 \text{ g}\]

Apply percent yield

Given that the actual reaction yield is 76%, we calculate the actual mass of \(\mathrm{CH}_2\mathrm{Cl}_2\) produced:\[\text{Actual mass} = 264.71 \text{ g} \times 0.76 = 201.18 \text{ g}\]

Key concepts

Chemical Reactions

A chemical reaction involves the transformation of substances through breaking and forming of chemical bonds. When methane (\(\mathrm{CH}_4\)) reacts with chlorine (\(\mathrm{Cl}_2\)), it leads to the formation of new substances: dichloromethane (\(\mathrm{CH}_2\mathrm{Cl}_2\)) and hydrogen chloride (\(\mathrm{HCl}\)).

• The initial substances (reactants) used here are methane and chlorine.
• The products, what are produced as a result of the reaction, are dichloromethane and hydrogen chloride.

Understanding this transformation is key to stoichiometry, which is the area of chemistry focused on the quantitative relationships between reactants and products in chemical reactions.The chemical reaction can be visually represented in a chemical equation, which shows the reactants changing into products.

Molar Mass Calculations

Molar mass is the mass of one mole of a substance. It is a crucial component of stoichiometry as it allows us to convert between the mass of a substance and the amount in moles.To calculate molar mass:

• Add the atomic masses of all atoms in a molecule. For example, methane (\(\mathrm{CH}_4\)) consists of one carbon atom (12.01 g/mol) and four hydrogen atoms (each 1.01 g/mol), totaling approximately 16.05 g/mol.
• This unit, grams per mole, helps us relate the mass of a sample to the number of molecules it contains.

In the exercise, the molar mass of \(\mathrm{Cl}_2\) is 70.90 g/mol. Calculating how many grams or moles are needed is essential to moving forward with solving how much product will be formed.

Percent Yield

Percent yield tells us how efficient a chemical reaction is in producing the desired product. In almost all practical scenarios, reactions yield less than 100% due to various factors such as incomplete reactions or side reactions.

• It is calculated using the formula: \(\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100\% \)
• In this exercise, with a theoretical yield of 264.71 grams of \(\mathrm{CH}_2\mathrm{Cl}_2\), the actual yield, considering a 76% efficiency, is 201.18 grams.

Understanding percent yield lets us estimate the actual results from a chemical reaction under non-ideal conditions.

Balancing Chemical Equations

Balancing chemical equations is a vital skill in chemistry, ensuring that the same number of each kind of atom appears on both sides of the equation.The exercise illustrates this by balancing the reaction:\[\mathrm{CH}_4 + 2\mathrm{Cl}_2 \rightarrow \mathrm{CH}_2\mathrm{Cl}_2 + 2\mathrm{HCl}\]

• First, count the number of each type of atom on both sides of the equation.
• Adjust coefficients (the numbers in front of compounds) to make sure each type of atom has the same count on either side.

Knowing how to properly balance equations is crucial in stoichiometry, as it allows you to use the correct mole ratios in calculations, ensuring precise results pertaining to reactants and products.