This is a limiting reagent problem and you need to identify which reagent is the limiting reagent (LR). I know it is a LR problem because the amounts of BOTH reactants are given. I solve these by solving two simple stoichiometry problems (simple meaning they aren't limiting reagent problems). First use one reagent and find moles of the product. Second, use the other reagent and find moles of the product. The two answers probably will differ; the correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that smaller value is the limiting reagent. Here is an example problem of a simple stoichiometry problem.
http://www.jiskha.com/science/chemistry/stoichiometry.html
After you know the LR, use the coefficients between the LR and the other reagent to calculate the number of moles of that reagent used. subtract from the original amount, the convert to grams.
Ammonium sulfate reacts with a solution of sodium hydroxide to produce sodium sulfate, ammonia gas and water. Calculate the yield in grams of ammonia gas that would form; and the amount of excess reagent in grams that would be left in the flask, after 9.92 grams of ammonium sulfate reacts with 250mL of 0.5 M sodium hydroxide
1 answer
1 answer