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Question: For the balanced equation 2C6H6O2 + 13O2 --> 12CO2 + 6H2O, if the reaction of 87.3 grams of O2 produces 7.54 grams of H2O, 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 O2
Step Two: Find the theoretical yield.
Question: For the balanced equation 2C6H6O2 + 13O2 --> 12CO2 + 6H2O, if the reaction of 87.3 grams of O2 produces 7.54 grams of H2O, 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 O2
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 O2 and the balanced equation is 2C6H6O2 + 13O2 --> 12CO2 + 6H2O, the ratio between O2 and H2O is a 13:6 or a 2 1/6:1 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 2.728 moles of O2 (^number of moles of O2 that we have^) we need six thirteenths (6/13) as many moles of H2O, as shown in the ratio, therefore, we divide the number of moles of O2 by 13/6 to get 1.259 moles of H2O.
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 H2O: H: (2x1) + O: (1x16) = 18 GFW grams=(1.259moles)(18GFW) ---> grams=22.662g
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 O2 and the balanced equation is 2C6H6O2 + 13O2 --> 12CO2 + 6H2O, the ratio between O2 and H2O is a 13:6 or a 2 1/6:1 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 2.728 moles of O2 (^number of moles of O2 that we have^) we need six thirteenths (6/13) as many moles of H2O, as shown in the ratio, therefore, we divide the number of moles of O2 by 13/6 to get 1.259 moles of H2O.
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 H2O: H: (2x1) + O: (1x16) = 18 GFW grams=(1.259moles)(18GFW) ---> grams=22.662g
Step Three: Plug in known information into equation shown below
Percentage Yield = (Actual Yield/Theoretical Yield) x 100%
Percentage Yield = (7.54/22.662) x 100%
Percentage Yield = 33.3%
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Percentage Yield = (Actual Yield/Theoretical Yield) x 100%
Percentage Yield = (7.54/22.662) x 100%
Percentage Yield = 33.3%
click here to go back to the quiz