# If the reaction of 91.3 grams of C3H6 produces a 81.3% yield, how many grams of CO2 would be produced? 2C3H6+9O2=>6CO2+6H2O

Dec 12, 2014

The answer is $236 g$.

Starting from the balanced chemical equation

$2 {C}_{3} {H}_{6} + 9 {O}_{2} \to 6 C {O}_{2} + 6 {H}_{2} O$

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

Now, since the reaction's percent yield is 81.3%, we can say that not all ${C}_{3} {H}_{6}$ was used in the reaction <=> ${O}_{2}$ is the limiting reagent.

The number of ${C}_{3} {H}_{6}$ moles, knowing its molar mass is $42 \frac{g}{m o l}$, is

${n}_{{C}_{3} {H}_{6}} = {m}_{{C}_{3} {H}_{6}} / \left(m o l a r m a s s\right) = \frac{91.3 g}{42 \frac{g}{m o l}} = 2.2$ moles.

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

${n}_{C {O}_{2}} = {n}_{{C}_{3} {H}_{6}} \cdot 3 = 6.6$ moles, and ${m}_{C {O}_{2}} = {n}_{C {O}_{2}} \cdot \left(m o l a r m a s s\right) = 6.6 m o l e s \cdot 44 \frac{g}{m o l} = 290 g$

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

%yield = n_(actual)/n_(the o re tic) * 100% ->

n_(actual) = (%yield * n_(the o retic))/(100%) = (81.3% * 6.6)/(100%) = 5.37 moles

Therefore, the actual mass of $C {O}_{2}$ produced is

${m}_{C {O}_{2}} = {n}_{a c t u a l} \cdot 44 \frac{g}{m o l} = 5.37 m o l e s \cdot 44 \frac{g}{m o l} = 236 g$