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marishachu [46]
3 years ago
11

he equilibrium constant, Kc , for the following reaction is 7.00×10-5 at 673 K.NH4I(s) NH3(g) + HI(g)If an equilibrium mixture o

f the three compounds in a 4.90 L container at 673 K contains 1.45 mol of NH4I(s) and 0.405 mol of NH3, the number of moles of HI present is moles.
Chemistry
1 answer:
frosja888 [35]3 years ago
3 0

<u>Answer:</u> The number of moles of HI in the solution is 1.24\times 10{-3} moles.

<u>Explanation:</u>

We are given:

K_c=7.00\times 10^{-5}\\n_{NH_3}=0.405mol\\n_{NH_4I}=1.45mol\\V=4.90L

To calculate the concentration of a substance, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     ......(1)

  • Concentration of ammonia:

[NH_3]=\frac{0.405mol}{4.90L}=0.083mol/L

  • Concentration of ammonium iodide:

[NH_4I]=\frac{1.45mol}{4.90L}=0.30mol/L

For the given chemical reaction:

NH_4I(s)\rightleftharpoons NH_3(g)+HI(g)

The expression of K_c for above equation follows:

K_c=\frac{[HI][NH_3]}{[NH_4I]}

Putting values in above equation, we get:

7.0\times 10^{-5}=\frac{[HI]\times 0.083}{0.30}

[HI]=2.53\times 10^{-4}

Calculating the moles of hydrogen iodide by using equation 1, we get:

2.53\times 10^{-5}=\frac{\text{Moles of HI}}{4.9}\\\\\text{Moles of HI}=1.24\times 10^{-3}

Hence, the number of moles of HI in the solution is 1.24\times 10{-3} moles.

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