Question

Calculate the number of grams of HI that are at equilibrium with 1.25 mol of H2 and 63.5 g of iodine at 448 °C.

a) 126.75 g
b) 63.38 g
c) 189.13 g
d) 94.56 g

Answer 1

To solve this problem, convert the amount of H2 and I2 given into moles. Apply the stoichiometry of the balanced equation to determine the number of moles of HI that will be at equilibrium. Finally, convert the moles of HI to grams.

Step-by-step explanation:

To solve this problem, we can use the equation and equilibrium constant given:

H₂(g) + I₂(g) → 2 HI(g)

Kc = 50.2

First, we need to convert the amount of H₂ and I₂ given into moles. H₂ has 1.25 mol and I₂ has 63.5 g. Converting the grams of I₂ to moles (using the molar mass of iodine) gives us 0.25 mol. Next, we need to apply the stoichiometry of the balanced equation to determine the number of moles of HI that will be at equilibrium.

Since the stoichiometric coefficient of HI is 2, we multiply the number of moles of I₂ by 2. This gives us 0.50 mol. Finally, we can convert the moles of HI to grams by multiplying by the molar mass of HI (127.91 g/mol). The result is 63.95 g, which is closest to option b) 63.38 g.