Question

For the following reaction, 4.34 grams of sulfur dioxide are mixed with excess oxygen gas . The reaction yields 3.89 grams of sulfur trioxide .
sulfur dioxide ( g ) + oxygen ( g ) sulfur trioxide ( g )

What is the theoretical yield of sulfur trioxide ?
grams
What is the percent yield for this reaction ?

Answer 1

Answer:

5.42g, 71.77%

Explanation:

First, we have to write out the balanced chemical equation. The unbalanced equation can be written as “SO2+O2 -> SO3” and to balance it, we can see that having two mols of SO2 and two mols of SO3 will make each side have the same amount of mols per element on each side. So the balanced chemical equation is “2SO2 + O2 -> 2SO3”

Now, we want to solve for the theoretical yield in grams of SO3. To do this, we have to use dimensional analysis. We convert g SO2 into mols SO2 using the molar mass of the elements. Then we convert mols of SO2 into mols of SO3 using the balanced equation. Once we’ve done that, we can convert mols of SO3 into grams of SO3.

You should know how to look up the molar mass of elements on the periodic table by now. Find the masses and set up the terms so they cancel like so:

$4.34g \times \frac{1mol \: so2}{64.07g \: so2} \times \frac{2 \: mol \: so3}{2 \: mol \: so2} \times \frac{80.07gso3}{1 \: mol \: so3}$

Doing the math, we get 5.42g so3 as the theoretical yield. This is the most amount that you could ever get if the world was a perfect place. But alas, it isn’t and mistakes are gonna happen, so the number is going to be less than that. So the best we can do, is to figure out the percent yield that we got.

In a lab scenario, this was calculated to be 3.89 g as stated by the problem. The percent composition formula is

$| \frac{result}{expected \: result} | \times 100$

and plugging the numbers into it, we get:

$| \frac{3.89}{5.42} | \times 100 = 71.77\%$

make sure to follow the decimal/significant figure rules of your instructor, but only round at the end. My professor didn't care too much thankfully, but some professors do