CHM 2045 Lecture Notes - Lecture 2: Potassium Bromide, Nonmetal, Ion

• School: University of Florida
• Department: Chemistry
• Course: CHM 2045
• Professor: Korolev Maria

Document Summary

We must go through several steps using the given values before we can actually figure out how much d can be prepared. A is the limiting reactant and b is the excess reactant. This simplifies the problem a bit, as equation 1 already begins with 2 mol of a and we can tell that 2 mol of a will make 4 mol of c. However, you must remember we are told equation 1 is only 75% efficient, so the actual yield of equation 1 will be 3 mol c [(4 mol c x 0. 75)= 3 mol c] Now we can use the amount of c to find the maximum amount of d we can have prepared: 3 mol c x (2 mol d/3 mol c) = 2 mol d. This works out well like with equation 1, and we know that 2 mol of d can be made with 3 mol of c.