# 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

Dec 13, 2014

The answer is $6.7 g$.

Starting from the balanced chemical equation

${C}_{6} {H}_{6} {O}_{3} + 6 {O}_{2} \to 6 C {O}_{2} + 3 {H}_{2} O$

we can see that we have a $1 : 3$ mole ratio between ${C}_{6} {H}_{6} {O}_{3}$ and ${H}_{2} O$; that is, for every mole of ${C}_{6} {H}_{6} {O}_{3}$ used in the reaction, $3$ moles of ${H}_{2} O$ are produced.

SInce the reaction's percent yield is 39.0%, we can say that not all ${C}_{6} {H}_{6} {O}_{3}$ was used in the reaction <=> ${O}_{2}$ is the limiting reagent.

The number of ${C}_{6} {H}_{6} {O}_{3}$ moles, knowing its molar mass is $126 \frac{g}{m o l}$, is

${n}_{{C}_{6} {H}_{6} {O}_{3}} = {m}_{{C}_{6} {H}_{6} {O}_{3}} / \left(m o l a r m a s s\right) = \frac{40.8 g}{126 \frac{g}{m o l}} = 0.32$ moles

Now, if all the ${C}_{6} {H}_{6} {O}_{3}$ would have reacted, the number of moles and, subsequently, the mass of ${H}_{2} O$ produced would have been

${n}_{{H}_{2} O} = 3 \cdot {n}_{{C}_{6} {H}_{6} {O}_{3}} = 3 \cdot 0.32 = 0.96$ moles, and

${m}_{{H}_{2} O} = {n}_{{H}_{2} O} \cdot \left(m o l a r m a s s\right) = 0.96 m o l e s \cdot 18 \frac{g}{m o l} = 17.3 g$

However, we know that the actual number of moles of ${H}_{2} O$ 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 ${H}_{2} O$ produced is

${m}_{{H}_{2} O} = {n}_{a c t u a l} \cdot \left(m o l a r m a s s\right) = 0.37 m o l e s \cdot 18 \frac{g}{m o l} = 6.7 g$