If the reaction of 40.8 grams of C6H6O3 produces a 39.0% yield, how many grams of H2O would be produced ? C6H6O3+6O2=>6CO2+3H2O

Answer 1
The answer is #6.7g#.

Initially focusing on the chemical equation in balance

#C_6H_6O_3 + 6O_2 -> 6CO_2 + 3H_2O#
we can see that we have a #1:3# mole ratio between #C_6H_6O_3# and #H_2O#; that is, for every mole of #C_6H_6O_3# used in the reaction, #3# moles of #H_2O# are produced.
SInce the reaction's percent yield is #39.0%#, we can say that not all #C_6H_6O_3# was used in the reaction <=> #O_2# is the limiting reagent.
The number of #C_6H_6O_3# moles, knowing its molar mass is #126g/(mol)#, is
#n_(C_6H_6O_3) = m_(C_6H_6O_3)/(molarmass) = (40.8g)/(126g/(mol)) = 0.32# moles
Now, if all the #C_6H_6O_3# would have reacted, the number of moles and, subsequently, the mass of #H_2O# produced would have been
#n_(H_2O) = 3 * n_(C_6H_6O_3) = 3 * 0.32 = 0.96# moles, and
#m_(H_2O) = n_(H_2O) * (molarmass) = 0.96 mol es * 18g/(mol) = 17.3g#
However, we know that the actual number of moles of #H_2O# produced is
#%yield = n_(actual)/n_(theo retic) * (100%) ->#
#n_(actual) = (%yield * n_(theo retic))/(100%) = (39.0 * 0.96)/(100%) = 0.37# moles
Therefore, the mass of #H_2O# produced is
#m_(H_2O) = n_(actual) * (molarmass) = 0.37 mol es * 18g/(mol) = 6.7g#
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Answer 2

First, calculate the molar mass of C6H6O3, which is 126.11 g/mol. Then, convert the given mass of C6H6O3 to moles. Next, use the stoichiometric ratio from the balanced equation to find the moles of H2O produced. Finally, convert the moles of H2O to grams using the molar mass of H2O, which is 18.015 g/mol.

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Answer 3

First, calculate the theoretical yield of H2O using the given mass of C6H6O3 (40.8 grams) and the stoichiometry of the reaction.

1 mole of C6H6O3 produces 3 moles of H2O.

Next, determine the number of moles of C6H6O3 using its molar mass. Then, use the stoichiometric ratio to find the number of moles of H2O produced.

Once you have the number of moles of H2O, use its molar mass to find the mass of H2O produced.

Finally, apply the percent yield to find the actual yield of H2O.

Calculate:

  1. Moles of C6H6O3
  2. Moles of H2O
  3. Mass of H2O
  4. Actual yield of H2O.
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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