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.

511.6 g is the theoretical yield in grams and 25%, the percent yield for this reaction

Explanation:

The reaction is:

H₂ (g)  +  I₂ (g)  → 2HI (g)

We have the amount of each reactant:

2 moles of I₂

3 moles of H₂

As ratio is 1:1, If I have 3 moles of hydrogen I may need 3 moles of iodine; the thing is I have 2 moles therefore the I₂ is my limiting reactant.

To determine the theoretical yield I make this rule of three:

1 mol of Iodine can produce 2 moles of iodide

Therefore 2 moles of Iodine may produce (2 .2)/1 = 4 moles of HI

We convert the moles to mass, to find the theoretical yield in grams:

4 mol . 127.9g / 1mol = 511.6 g

To find the percent yield we follow this formula:

(Yield produced/ Theoretical yield) . 100 = (1 mol/ 4mol) .100 = 0.25

the answer is d. 74.02