Question

In a laboratory experiment, the reaction of 3.0 mol of H₂ with 2.0 mol of I₂ produced 1.0 mol of HI. Determine the theoretical yield in grams and the percent yield for this reaction.

a) Theoretical yield: 151 g, Percent yield: 66.7%
b) Theoretical yield: 99 g, Percent yield: 44.0%
c) Theoretical yield: 202 g, Percent yield: 89.3%
d) Theoretical yield: 121 g, Percent yield: 53.3%

Answer 1

Iodine (I₂) is the limiting reactant, yielding a theoretical yield of 515.672 g of HI. With an actual yield of 128.918 g of HI, the percent yield is 25.0%. The student's provided options do not match this calculation.

Step-by-step explanation:

Theoretical and Percent Yield Calculation

Let's first determine the theoretical yield for the reaction H₂(g) + I₂(g) → 2 HI(g), given that we start with 3.0 mol of H₂ and 2.0 mol of I₂.

First, look at the stoichiometry of the reaction:

1. 1 mol of H₂ reacts with 1 mol of I₂ to produce 2 mol of HI.
2. Therefore, 3.0 mol of H₂ would ideally produce 3.0 x 2 = 6.0 mol of HI, if H₂ were the limiting reactant.
3. However, only 2.0 mol of I₂ are available, which would produce 2.0 x 2 = 4.0 mol of HI, if I₂ were the limiting reactant.

Since 4.0 mol of HI is less than 6.0 mol, I₂ is the limiting reactant, and the theoretical yield is 4.0 mol of HI. To convert moles to grams, multiply by the molar mass of HI (127.91 g/mol for I + 1.008 g/mol for H = 128.918 g/mol for HI):

Theoretical yield = 4.0 mol x 128.918 g/mol = 515.672 g HI

However, only 1.0 mol of HI was produced, so:

Actual yield = 1.0 mol x 128.918 g/mol = 128.918 g HI

To find the percent yield, we use the formula:

Percent Yield = (Actual Yield/Theoretical Yield) x 100%

Percent Yield = (128.918 g / 515.672 g) x 100% = 25.0%

None of the provided answers match this calculation, so it seems that there might be an error in the given answer choices or the student's reaction data.