181k views
2 votes
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!)​

User Kudzu
by
5.0k points

1 Answer

2 votes

Answer:

amount of H₂O = 6.825 g

Step-by-step 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

User Kevin Brotcke
by
4.4k points