Calculate the number of moles of HI that are at equilibrium with 1.25 mol of H2 and 1.25 mol of I2 in a 5.00-L flask at 448 °C. H2 + I2 ⇌ 2HI Kc = 50.2 at 448 °C
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25 mol of H2 and 1.25 mol of I2, and no HI. Let's say that x mol of H2 and I2 react to form 2x mol of HI. At equilibrium, we have (1.25 - x) mol of H2, (1.25 - x) mol of I2, and 2x mol of HI. The equilibrium concentrations are then (1.25 - x) / 5.00 M for H2 Show more…
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