Setting up the Keq expression:

$\overline{){\mathbf{Keq}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}\frac{\left[\mathrm{products}\right]}{\left[\mathrm{reactants}\right]}}\phantom{\rule{0ex}{0ex}}\mathbf{50}\mathbf{.}\mathbf{2}\mathbf{}\mathbf{=}\mathbf{}\frac{{\left({\displaystyle \frac{\mathrm{moles}\mathrm{HI}}{5L}}\right)}^{2}}{\left({\displaystyle \frac{1.25\mathrm{mol}{I}_{2}}{5L}}\right)\left({\displaystyle \frac{1.25\mathrm{mol}{H}_{2}}{5L}}\right)}\phantom{\rule{0ex}{0ex}}\frac{\mathbf{moles}\mathbf{}\mathbf{HI}}{\mathbf{5}\mathbf{}\mathbf{L}}\mathbf{=}\mathbf{}\sqrt{\mathbf{(}\mathbf{50}\mathbf{.}\mathbf{2}\mathbf{)}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{I}}_{\mathbf{2}}\mathbf{)}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{)}}\phantom{\rule{0ex}{0ex}}\mathbf{moles}\mathbf{}\mathbf{HI}\mathbf{}\mathbf{=}\mathbf{}(1.771)\left(5\right)$

Calculate the number of moles of HI that are at equilibrium with 1.25 mol of H_{2} and 1.25 mol of I_{2} in a 5.00−L flask at 448°C.

H_{2} + I_{2} ⇌ 2HI K _{c} = 50.2 at 448°C

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