Here is an example of a stoichiometry problem I've posted. Just follow the procedure shown to work the second part of your question. For the first part, see how much AlCl3 (with excess Al) can be produced with 36.0 g Cl2, compare with the answer from the second part. The value for both, where one functions as a limiting reagent, will be the smaller of the two.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Please help solve;
2Al +3Cl2 --> 2AlCl3
You are give 31.0g Al and 36.0g of Cl2.
If you had excess chlorine, how many moles of of aluminum chloride could be produced from 31.0 g of aluminum?
2 answers
figure the moles you have of aluminum, and chlorine.
Start with the aluminum. IF you have more than 3/2 of that of chlorine, then aluminum is the limiting reageant. If you have less than 3/2 of chlorine, chlorine is the limiiting reageant.
Take the mole of the limiting regeant, and use the mole ratio in the balanced equation to gind moles of aluminum cloride.
Example: aluminum is the limiting regent. If you have 3.4 moles of aluminum, you will get 2/2 * 3.4 moles of aluminum chloride.
Example: chlorine is the limiting regent. If you have 4.5 moles of chlorine, then you get 4.5*2/3 moles of aluminum cloride.
Start with the aluminum. IF you have more than 3/2 of that of chlorine, then aluminum is the limiting reageant. If you have less than 3/2 of chlorine, chlorine is the limiiting reageant.
Take the mole of the limiting regeant, and use the mole ratio in the balanced equation to gind moles of aluminum cloride.
Example: aluminum is the limiting regent. If you have 3.4 moles of aluminum, you will get 2/2 * 3.4 moles of aluminum chloride.
Example: chlorine is the limiting regent. If you have 4.5 moles of chlorine, then you get 4.5*2/3 moles of aluminum cloride.
0 answers