Question

If the reaction of 125 grams of C6H6O₃ reactsin excess of oxygen (O₂) and produces 51 gram ofH2O, What is the percentage yeild?C6 H6O3 +3O2 gives 6CO + 3H₂Oo) 80% 6) 85% C)101% d) 95%​

Answer 1

Equation For Reaction:

C6H6O3 + 3O2 -------> 6CO + 3H₂O

Given:

=> 125 grams of C6H6O3 burns in excess Oxygen to give 51 grams of Water.

To find:

1) Grams of Water that should have been formed if the reaction was completed & hence,

2) Percentage yield.

Let us find,

Molar mass of C6H6O3 = 6*12 + 6*1 + 3*16 =>72+6+48 => 126 g per mole.

So, 125 grams is 125/126 moles => 0.992 moles of C6H6O3.

=> From the equation, we can see that, 1 part of C6H6O3 reacts with 3 parts of Oxygen gas to give 6 parts of Carbon Monoxide and 3 parts of water.

So, 1 part = 0.992 moles

Hence, 3 parts of water is 0.992*3 => 2.976 moles of H2O.

Molar mass of H2O = 2*1 + 1*16 => 18 grams per mole.

So, 2.976 moles of H2O has a mass of 2.976*18 => 53.568 grams

So, mass of water that should have been formed = 53.568 g

Mass of water formed = 51 grams.

=> $Percentage Yield = \frac{Mass of product}{Expected mass} *100$

=> Percentage yield = (51/53.568)*100 => 95.20 % (approx.)

So, the percentage yield is (d) 95 % (approx.)