4. Write the balanced equation for CH4+Cl2„_CH2Cl2+HCl.
5. Using the equation in problem 4 if 6.00 g of CH4 are mixed with 6.00 g of CL2, how much HCl can be made? What is the percent yield if 2.5 g are recovered from the reaction in the lab?
CH4 + 2Cl2 ==> CH2Cl2 + 2HCl
This is a limiting reagent problem. Remember how do to this type problem. Most are alike.
2a. Convert 6.00 g CH4 to moles. moles = grams/molar mass.
2b. Convert 6.00 g Cl2 to moles. Same formula.
3a. Using the coefficients in the balanced equation, convert moles CH4 to moles HCl.
3b. Same process, convert moles Cl2 to moles HCl.
3c. It is quite likely that the answer to 3a and 3b will not be the same. Obviously, one of them is wrong; the correct answer is ALWAYS the smaller value.
4. Convert the value from 3c into grams. grams HCl = moles HCl x molar mass HCl. This is the theoretical yield. This is the end of the limiting reagent problem. Almost all of them can be worked with this four-step procedure.
5. percent yield = (actual yield/theoretical yield)*100 = ??
Actual yield is given in the problem as 2.5 g.
To write the balanced equation for the reaction between CH4 (methane) and Cl2 (chlorine) to produce CH2Cl2 (dichloromethane) and HCl (hydrochloric acid), follow these steps:
Step 1: Write the molecular formulas for the reactants and products:
Reactants: CH4 + Cl2
Products: CH2Cl2 + HCl
Step 2: Identify the different elements present in the reactants and products:
Reactants: C (carbon), H (hydrogen), Cl (chlorine)
Products: C (carbon), H (hydrogen), Cl (chlorine)
Step 3: Count the number of each type of atom on both sides of the equation and ensure they are equal:
Reactants: 1 carbon (C), 4 hydrogen (H), 2 chlorine (Cl)
Products: 1 carbon (C), 2 hydrogen (H), 3 chlorine (Cl)
Step 4: Balance the equation by adding coefficients to the compounds as needed:
CH4 + Cl2 -> CH2Cl2 + HCl
Now that we have the balanced equation, we can proceed to the next question.
To determine how much HCl can be made when 6.00 g of CH4 are mixed with 6.00 g of Cl2, we need to follow these steps:
Step 1: Convert the given masses of CH4 and Cl2 to moles using their molar masses:
Molar mass of CH4 (methane) = 12.01 g/mol (C) + 4 * 1.008 g/mol (H) = 16.04 g/mol
Molar mass of Cl2 (chlorine) = 35.45 g/mol (Cl) + 35.45 g/mol (Cl) = 70.90 g/mol
Moles of CH4 = 6.00 g / 16.04 g/mol = 0.374 mol
Moles of Cl2 = 6.00 g / 70.90 g/mol = 0.0844 mol
Step 2: Determine the limiting reactant:
To determine the limiting reactant, we compare the moles of each reactant to their stoichiometric coefficients in the balanced equation. The reactant with fewer moles compared to its coefficient is the limiting reactant.
From the balanced equation, 1 mole of CH4 reacts with 1 mole of Cl2 to produce 1 mole of CH2Cl2 and 1 mole of HCl.
Since the ratio of CH4 to Cl2 is 1:1, and we have 0.374 mol of CH4 and 0.0844 mol of Cl2, the limiting reactant is Cl2.
Step 3: Calculate the moles of HCl produced from the limiting reactant:
From the balanced equation, we know that 1 mole of Cl2 produces 1 mole of HCl.
Therefore, moles of HCl = moles of Cl2 = 0.0844 mol
Step 4: Convert moles of HCl to grams using its molar mass:
Molar mass of HCl (hydrochloric acid) = 1.008 g/mol (H) + 35.45 g/mol (Cl) = 36.46 g/mol
Mass of HCl = moles of HCl * molar mass of HCl = 0.0844 mol * 36.46 g/mol = 3.07 g
So, when 6.00 g of CH4 are mixed with 6.00 g of Cl2, it will produce approximately 3.07 g of HCl.
To calculate the percent yield of HCl, given that 2.5 g are recovered from the reaction in the lab, follow these steps:
Step 1: Calculate the percent yield:
Percent Yield = (Actual yield / Theoretical yield) * 100
Theoretical yield is the calculated mass of HCl based on the limiting reactant, which we calculated to be 3.07 g.
Actual yield = 2.5 g
Percent Yield = (2.5 g / 3.07 g) * 100 = 81.5%
Therefore, the percent yield of HCl in this reaction is 81.5%.