Chapter 5, Problem 89P

The local anesthetic ethyl chloride ${\text{(C}}_{\text{2}}{\text{H}}_{\text{5}}\text{Cl, molar mass 64}\text{.51 g/mol)}$ can be prepared by reaction of ethylene ${\text{(C}}_{\text{2}}{\text{H}}_{\text{4}}\text{, molar mass 28}\text{.05 g/mol)}$ with HCl $\text{(molar mass 36}\text{.46 g/mol),}$ according to the balanced equation, ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}+\text{HCl}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{Cl}$ .

1. If 8.00 g of the ethylene and 12.0 g of HCl are used, how many moles of each reactant are used?
2. What is the limiting reactant?
3. How many moles of product are formed?
4. How many grams of product are formed?
5. If 10.6 g of product are formed, what is the percent yield of the reaction?