2Na + 2HCl --> 2NaCl + H2
If you start with 5g Na and 5g HCl, which is the limiting reactant? How much of the excess reactant is left at the end of the reaction?
Explanation
Answer: E. Solution is as follows:
5g Na * 1 mol Na/22.99g * 2 mol NaCl/2 mol Na * 58.44g NaCl/1 mol NaCl = 12.7 g NaCl
5g HCl * 1mol HCl/36.46g * 2mol NaCl/2mol HCl * 58.44g NaCl/1 mol NaCl = 8.01 g NaCl
So, HCl is the limiting reactant because it produces less product. The next step is to determine the excess reactant. Excess reactant comes from the non-limiting reactant, which is sodium.
5g Hcl * 1mol HCl/36.46g *2mol Na/2mol HCl * 22.99g Na/1 mol = 3.15g
**This is the amount of sodium used in the reaction. To find the excess, subtract the amount used from the starting amount: 5g - 3.15g = 1.85g excess
Note: This is stoichiometry. The problem can be solved by relating it to the amount of H2 gas produced, too, instead of using NaCl. You will get the same answer.
