Most limiting reagent problems are worked the same way. Print these instructions and memorize them.
1. Write and balanced the equation. You have that.
2. USUALLY you are given grams and you must convert to moles by moles = grams/molar mass. This already has the moles given and you can skip this step.
3a. Using the coefficients in the balanced equation, convert 3.0 moles Al to moles of AlCl3.
3b. Same procedure but convert 5.3 moles Cl2 to moles AlCl3.
3c. It is quite likely that the answer to 3a and 3b will not be the same which means one of them is wrong. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
4. USUALLY, the problem asks for grams here and you would calculate grams AlCl3 from the smaller value = moles x molar mass. That isn't needed here since they don't ask for it but this is useful information for the future.
5. To do part d, you have identified the limiting reagent. The other reagent is not limiting; therefore, there will be some that remains unreacted. To know the final value, use the coefficients (as in step 3 above) to determine the moles that react with ALL of the limiting reagent, subtract from the initial amount, the different is the amount remaining unreacted. If you wish (although the problem doesn't ask for it), you can multiply moles remaining x molar mass to arrive at the grams of the other reactant after the reaction.
In a reaction chamber, 3.0mol of aluminum is mixed with 5.3mol Cl2 and reacts. The reaction is described by the following balanced chemical equation.
2Al + 3Cl2 ==> 2AlCl3
a)Identify the limiting reagent for the reaction.
b)Calculate the number of moles or products formed
c)Calculate the number of moles of excess reagent remaining after the reaction.
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