4. Write the balanced equation for CH4+Cl2„_CH2Cl2+HCl.

Question

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?

Answer 1

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.