Respuesta :
Answer: The number of moles of HI in the solution is [tex]1.24\times 10{-3}[/tex] moles.
Explanation:
We are given:
[tex]K_c=7.00\times 10^{-5}\\n_{NH_3}=0.405mol\\n_{NH_4I}=1.45mol\\V=4.90L[/tex]
To calculate the concentration of a substance, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex] ......(1)
- Concentration of ammonia:
[tex][NH_3]=\frac{0.405mol}{4.90L}=0.083mol/L[/tex]
- Concentration of ammonium iodide:
[tex][NH_4I]=\frac{1.45mol}{4.90L}=0.30mol/L[/tex]
For the given chemical reaction:
[tex]NH_4I(s)\rightleftharpoons NH_3(g)+HI(g)[/tex]
The expression of [tex]K_c[/tex] for above equation follows:
[tex]K_c=\frac{[HI][NH_3]}{[NH_4I]}[/tex]
Putting values in above equation, we get:
[tex]7.0\times 10^{-5}=\frac{[HI]\times 0.083}{0.30}[/tex]
[tex][HI]=2.53\times 10^{-4}[/tex]
Calculating the moles of hydrogen iodide by using equation 1, we get:
[tex]2.53\times 10^{-5}=\frac{\text{Moles of HI}}{4.9}\\\\\text{Moles of HI}=1.24\times 10^{-3}[/tex]
Hence, the number of moles of HI in the solution is [tex]1.24\times 10{-3}[/tex] moles.
