Because CH4 is the limiting reagent, we must find how many moles we have. 1. 5.14 g CH4 * (1 mol CH4 / 16.04 g/mol) = .32125 moles CH4 Now we know that there are .32125 moles of CCl4 produced as well. Next, take .32125 moles and multiply it by the atomic mass of CCl4, which is 153.82 grams = 49.29 g CCl4 produced by the reaction.
What mass of CCl4 is formed by the reaction of 5.14 g of methane with an excess of chlorine?