50.0mL of 1.04moles HA is added to 50.0mL if 0.90moles NH4OH.With a 1:1 stoichiometric ratio. What is the limiting reactant in moles of this reaction.

Answer 1

The problem is confusing IF you have posted it correctly. Have you added 1.04 mols HA or have you added 50.0 mL of 1.04M HA? Same with NH4OH. Is that 0.90 moles or 50.0 mL 0.90 M NH4OH? Actually it doesn't make any difference. The LR is NH4OH with either scenario.

Answer 2

From you post above
DrBob- it says my answer should be a number.

Then it depends upon the answer to the first question I asked. I'll do it both ways. If you have 1.04 mols HA and 0.90 mols NH4OH (as your problem states), then,
.........HA + NH4OH ==> NH4A + H2O
........1.04...0.90.......
.......-0.90..-0.90.......
...0.14 excess...0
So the limiting reactant is 0.90 mol NH4OH before the reaction or zero mol NH4OH after the reaction. I can't tell from the question which is required.

If instead you have 50.0 mL of 1.04 M HA (that is 0.0500 x 1.04 = 0.0520 mols) mixed with 50.0 mL of 0.90 M NH4OH (that is 0.0500 x 0.90 M = 0.045), then
..........HA + NH4OH ==> NH4A + H2O
........0.0520...0.045
......-0.045...-0.045
......0.007xs.....0
Then the LR is 0.045 mols NH4OH before the reaction or zero mols NH4OH after the reaction. Ditto above, I can't tell from the question which is required.

Answer 3

THANK YOU!!!!