Answer to Question #64764 in General Chemistry for christian witmer

Question

) If you react 5.00 g of aluminum with 20.0 g of iodine, how much aluminum iodide do you expect to produce from this reaction? What is the limiting reactant? What is the excess reactant? How much of the excess reactant do you expect to have left over?

Answer 1

Solution:
1) Write the balanced equation:
2Al + 3I2 = 2AlI3
We can see that 2 moles of aluminum react with 3 moles of iodine giving 2 moles of aluminum iodide.
2) Convert moles to masses.
Molar mass of Al = 27.0 g/mol;
Molar mass of I2 = 126.9 * 2 = 253.8 g/mol;
Molar mass of AlI3 = 27.0 + 126.9 * 3 = 407.7 g/mol;
So (2 * 27) = 54.0 g of aluminum react with (3 * 253.8) = 761.4 g of iodine giving (2 * 407.7) = 815.4 g of aluminum iodide.
3) Find which reactant is in excess.
If 54.0 g of aluminum requires 761.4 g of iodine than 5.00 g of aluminum would require
5.00 * 761.4 / 54.0 = 70.5 g of iodine. In fact we have 20.0 g of iodine so iodine is in shortage. Iodine is the limiting reactant and aluminum is the excess reactant.
4) Find what we obtain.
The amount of final product is determined by the amount of limiting reactant.
761.4 g of iodine produce 815.4 g of aluminum iodide. So 20.0 g of iodine would produce 20.0 * 815.4 / 761.4 = 21.4 g of aluminum iodide.
761.4 g of iodine requires 54.0 g of aluminum for the reaction, so 20.0 g of iodine would require 20.0 * 54.0 / 761.4 = 1.4 g of aluminum. And if we have 5.00 g of aluminum than we would have (5.00 – 1.4) = 3.6 g left over.

Answer:
I expect from this reaction to produce 21.4 g of aluminum iodide
Iodine is the limiting reactant.
Aluminum is the excess reactant.
I expect to have 3.6 g of aluminum left over.