Chapter 5, Problem 90P

The solvent dischloromethane $({\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{, molar mass 84}\text{.93 g/mol})$ can be prepared by reaction of methan $({\text{CH}}_{\text{4}}\text{, molar mass 16}\text{.04 g/mol})$ with Cl₂ (molar mass 70.90 g/mol), according to the balanced equation, ${\text{CH}}_{\text{4}}+{\text{2Cl}}_{\text{2}}\to {\text{CH}}_{\text{2}}{\text{Cl}}_{\text{2}}\text{+2HCl}\text{.}$

1. If 5.00 g of methane and 15.0 g of Cl₂ 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 5.6 g of product are formed, what is the percent yield of the reaction?