For the following reaction, 4.41 grams of hydrogen gas are allowed to react with 39.8 grams of iodine . hydrogen(g) + iodine(s) hydrogen iodide(g)

Question

What is the FORMULA for the limiting reagent?

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

Answer 1

Answered previously

Formula of limiting reagent is I₂

How much H2 remains?

0.157 mols I2 x 1 mol H2 / mol I2 = 0.157 mols H2 used up

Started with 2.21 mols H2

mass H2 remaining = 2.21 mols - 0.157 mols = 2.053 mols x 2 g / mol = 4.11 g H2 remaining

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PREVIOUS WORK

Write the correctly balanced equation for the reaction:

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

Find the limiting reactant:

mols H₂ = 4.41 g H₂ x 1 mol H₂ / 2 g = 2.205 mols H₂

mols I₂ = 39.8 g x 1 mol I₂ / 254 g = 0.157 mols I₂

Since they react in a mol ratio of 1 : 1, the I₂ is the limiting reactant

Use mols of limiting reactant (0.157 mols I₂) to find theoretical yield of HI:

0.157 mols I₂ x 2 mols HI / mol I₂ x 128 g HI / mol HI = 40.2 g HI