5.)
for the balanced equation shown below, if the reaction of 40.8 grams of
c6h603 produces a 39.0% yield, how many grams of h2o would be produced? (hint
work backwards to find actual yield! )

amount of H₂O = 6.825 g

Explanation:

Data Given:

mass of C₆H₆O₃ = 40.8 g

percent yield = 39.0 %

amount of H₂O = ?

Solution:

First we have to write the complete reaction

             C₆H₆O₃ + 6 O₂ > 6 CO₂ + 3 H₂O

So, if we look at the reaction

             C₆H₆O₃ + 6 O₂ > 6 CO₂ + 3 H₂O

               1 mol                                              3 mol

From the above reaction we come to know that C₆H₆O₃ and H₂O are in 1 : 3 mole ration that is one mole of C₆H₆O₃ produce 3 moles of H₂O

As the percent yield is 39% so it means that all the C₆H₆O₃ has not used in the reaction and Oxygen is limiting reagent.

To Know the Theoretical yield of H₂O we have to convert moles to mass

molar mass of  C₆H₆O₃ = 126 g/mol

molar mass of H₂O = 18 g/mol

Now

               C₆H₆O₃       +      6 O₂ > 6 CO₂  +   3 H₂O

           1 mol  (126 g/mol)                                          3 mol (18 g/mol)

                  126 g                                                              54 g

Now come to know that if 126 g of C₆H₆O₃ produce  54 g of H₂O then wat mass of water will be produce by 40.8 g of C₆H₆O₃.

apply unity formula

              126 g of C₆H₆O₃  ≅ 54 g of H₂O

              40.8 g of C₆H₆O₃ ≅ X g of H₂O

By cross multiplication:

             mass of H₂O = 40.8 g x 54 g / 126 g

             mass of H₂O = 17.5 g

To find the actual mass of H₂O

formula will be used

         percent yield of H₂O = actual yield / theoretical yield x 100

Rearrange the above formula

      actual yield = (percent yield of H₂O / theoretical yield)/100 . . . . . (1)

Put values in equation 1

     actual yield of H₂O = (39 x  17.5 g ) x 100

     actual yield H₂O = (682.5 g ) x 100

     actual yield H₂O = 6.825 g

So Amount of H₂O  produce = 6.825 g

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