if you add 0.056 moles of HC2H3O2 to 0.064 moles of NH3, how many moles of each are actually neutralized?
HAc + NH3 ==> NH4Ac
0.056..0.064.....
Since these are 1:1 (1 mole HAc neutralizes exactly 1 mole NH3), then you can see that you can't react more than 0.056. So all of the HAc reacts and leaves 0.064-0.056 = 0.008 mole NH3 unreacted. How much NH4Ac is formed. 0.056 mole of course.
To determine how many moles of each compound are actually neutralized, we need to identify the stoichiometry of the reaction between HC2H3O2 (acetic acid) and NH3 (ammonia). The balanced chemical equation for the reaction between these two compounds is as follows:
HC2H3O2 + NH3 → NH4C2H3O2
From the balanced equation, we can see that the mole ratio between HC2H3O2 and NH3 is 1:1. This means that one mole of HC2H3O2 reacts with one mole of NH3.
Given that you have added 0.056 moles of HC2H3O2 and 0.064 moles of NH3, we can determine the limiting reactant. The limiting reactant is the one that gets completely consumed in the reaction, thereby determining the maximum amount of product formed.
To find the limiting reactant, we compare the moles of HC2H3O2 and NH3 added:
HC2H3O2: 0.056 moles
NH3: 0.064 moles
Since both compounds have a 1:1 stoichiometric ratio, the limiting reactant is the one with the lower number of moles. In this case, HC2H3O2 is the limiting reactant because it is present in a lower amount (0.056 moles compared to 0.064 moles of NH3).
Therefore, 0.056 moles of HC2H3O2 will react with 0.056 moles of NH3.
To determine the number of moles of each substance that are actually neutralized, we need to identify the balanced chemical equation for the reaction between HC2H3O2 (acetic acid) and NH3 (ammonia).
The balanced chemical equation for the reaction between acetic acid and ammonia is as follows:
HC2H3O2 + NH3 → NH4C2H3O2
From the equation, we can see that one mole of HC2H3O2 reacts with one mole of NH3 to produce one mole of NH4C2H3O2 (ammonium acetate). Therefore, the stoichiometry of the reaction is 1:1:1.
Given that you added 0.056 moles of HC2H3O2 and 0.064 moles of NH3, we can determine that the amount of HC2H3O2 that is actually neutralized will be equal to the smaller of the two amounts. In this case, it is 0.056 moles.
Since the stoichiometry is 1:1, the amount of NH3 that is actually neutralized will also be 0.056 moles.
Therefore, 0.056 moles of HC2H3O2 and 0.056 moles of NH3 are neutralized in the reaction.