## I'm sorry but the answer that you got was incorrect.

Let's go through the steps together to help you understand.

The limiting reagent is C2H4O2S

**Question:**For the balanced equation 2C2H4O2S + 7O2 --> 4CO2 + 4H2O + 2SO3, if the reaction of 42.0 grams of C2H4O2S produces a 32.4 grams of CO2, what is the percent yield?**Step One:**Identify the limiting reagent. (the question only gives you**one**of the reactants, therefore, making that compound the limiting reagent)The limiting reagent is C2H4O2S

**Step Two:**Find the theoretical yield.

*Here are the steps to finding the theoretical yield***Step One:**Use the limiting reagent and see what the ratio is between that and the compound in the product (this is what you are trying to figure out). Make this ratio 1:? or ?:1

Since the limiting reagent is C2H4O2S and the balanced equation is 2C2H4O2S + 7O2 --> 4CO2 + 4H2O + 2SO3, the ratio between C2H4O2S and CO2 is a 2:4 or a 1:2 ratio.

**Step Two:**Using the ratio between the limiting reagent and the compound in the product, find the number of moles in the product's compound by multiplying/diving the limiting reagent by the non-one number in the ratio (1:

**?**).

For every 0.467 moles of C2H4O2S (^number of moles of C2H4O2S that we have^) we need twice as many moles of CO2, as shown in the ratio, therefore, we multiply the number of moles of C2H4O2S by 2 to get 0.914 moles of CO2.

**Step Three:**Once you know how many moles of the compound in the product you have, turn the moles back into grams. This allows you to answer the question of how much of ___ compound is able to be produced.

To turn the moles back into grams we used the equation grams=(moles)(GFW).

GFW of CO2: C: (1x12) + O: (2x16) = 44 GFW grams=(0.914moles)(44GFW) ---> grams=40.216g

**Step Three:**Plug in known information into equation shown below

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

Percentage Yield = (32.4grams/40.216) x 100%

Percentage Yield = 80.7%

click here to go back to the quiz