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Question

For the following reaction, 0.372 grams of hydrogen gas are allowed to react with 50.1 grams of iodine.

hydrogen(g) + iodine(s)  hydrogen iodide(g)

What is the maximum mass of hydrogen iodide that can be formed?   grams

What is the FORMULA for the limiting reagent?

What mass of the excess reagent remains after the reaction is complete?   grams

Answer 1

H₂(g) + I₂(s) ==> 2HI(g) ... balanced equation

0.372g...50.1g.............

First, we must find which reactant is limiting:

0.372 g H₂ x 1 mol H₂/2 g = 0.186 mols H₂

50.1 g I₂ x 1 mol I₂ / 254 g = 0.197 mols I₂

Because they react in a 1:1 mol ratio, H2 will be limiting

FORMULA of limiting reagent = H₂(g)

MAXIMUM mass of HI that can form:

0.186 mols H₂ x 2 mols HI / mol H₂ x 128 g HI / mol HI = 47.6 g HI

Mass of excess reagent remaining:

mol I₂ used up = 0.186 mols H₂ x 1 mol I₂ / mol H₂ = 0.186 mols I₂ used

mols I₂ originally present = 0.197 mols (see above calculation)

mols I₂ remaining = 0.197 mols - 0.186 mols = 0.011 mols I₂ remaining

mass I₂ remaining = 0.011 mols I₂ x 254 g /mol = 2.79 g I₂ remaining