Fill in the following reaction table with millimoles (to the nearest 0.01 mmol)

Question

Fill in the following reaction table with millimoles (to the nearest 0.01 mmol) for the reaction occurring when 42.5 mL of 0.396-M ammonia are mixed with 66.6 mL of 0.374-M hydrofluoric acid.

What is the final concentration of ammonium ion to three significant figures?

[NH41+] =   M

How many mmols (three sig figs) of the limiting reactant are actually present at equilibrium? Note that this value is not based on the final amount shown in the reaction table because the reaction table assumes complete reaction (not an equilibrium) of the limiting reactant.

mmol

HF + NH3 F1- + NH41+ initial mmol delta mmol final mmol

Explanation / Answer

mmole = molarity x volume

mmole of NH3 = 0.396 x 42.5 = 16.83 mmol

mmole of HF = 0.374 x 66.6 = 24.91 mmol

This is a 1 : 1 ratio reaction, so the limiting reagent would be NH3

Table:

              HF +   NH3   <===>   F-    + NH4+

initial    24.91    16.83                0          0           mmol

delta    -16.83   -16.83             16.83   16.83      mmol

final        8.08        0                 16.83   16.83      mmol

Final concentration of [NH4+] = 16.830 mmol = 0.01683/0.0425+0.0666) = 0.154 M

mmols of limiting reactant present at equilibrium,

7 x 10^-4 = x^2/(0.23-x)

x = 0.0127 M

mmoles of limiting reactant present at equilibrium = 9.35 mmol

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