Answer:
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
Answer:
Explanation:
Given:
3.0 mol H2
2.0 mol I2
1.0 mol HI
First, balance the equation.
H2 + I2 --> 2HI
Next, determine the limiting reagent, and therefore you'll determine the theoretical yield of HI.
3 mol H2 × 2 mol HI / 1 mol H2 = 6 mol HI
2 mol I2 × 2 mol HI / 1 mol I2 = 4 mol HI
As I2 produced a smaller amount of HI, I2 is considered to be the limiting reagent, and 4 mol HI is considered to be the theoretical yield produced.
To answer the first part of the question, let's convert the 4 mol HI into grams
4 mol HI × 128 g HI / 1 mol HI = 512 g HI expected.
Now on to the second part of the question; we'll keep the units in "mol" when calculating the percent yield.
% = (actual yield / theoretical yield) × 100
The "actual yield" is the given amount of moles of the yielded product from the problem itself (sometimes the problem will give you this amount in grams, in which you'll have to convert to moles)
So,
% = (1.0 mol HI / 4 mol HI) × 100 = 25%
