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

Question

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

Owen Ambrose · Accepted Answer

The answer is #236g#. Initially focusing on the chemical equation in balance #2C_3H_6 + 9O_2 -> 6CO_2 + 6H_2O# we can see that we have a #2:6# (or a #1:3#) mole ratio between #C_3H_6# and #CO_2#; that is, for every mole of #C_3H_6# used in the reaction, #3# moles of #CO_2# are produced. Now, since the reaction's percent yield is #81.3%#, we can say that not all #C_3H_6# was used in the reaction <=> #O_2# is the limiting reagent.  The number of #C_3H_6# moles, knowing its molar mass is #42g/(mol)#, is #n_(C_3H_6) = m_(C_3H_6)/(molarmass) = (91.3g)/(42g/(mol)) = 2.2# moles.  If all the #C_3H_6# would have reacted, the number of moles and, subsequently, the mass of #CO_2# would have been #n_(CO_2) = n_(C_3H_6) * 3 = 6.6# moles, and #m_(CO_2) = n_(CO_2) * (molarmass) = 6.6 mol es * 44g/(mol) = 290g# However, we know that the actual number of #CO_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 #CO_2# produced is #m_(CO_2) = n_(actual) * 44g/(mol) = 5.37 mol es* 44 g/(mol) = 236g#

Luna Chen · Answer

undefined To calculate the grams of CO2 produced, you need to first find the moles of C3H6 used in the reaction, then determine the theoretical yield of CO2, and finally apply the given percentage yield to find the actual yield.

1. Calculate moles of C3H6:
   $91.3 \text{ g} \times \frac{1 \text{ mol}}{42.08 \text{ g/mol}} = 2.17 \text{ mol}$

2. Using the balanced equation, determine the theoretical yield of CO2:
   $2 \text{ mol C3H6} \times \frac{6 \text{ mol CO2}}{2 \text{ mol C3H6}} = 6 \text{ mol CO2}$

3. Calculate the actual yield using the given percentage yield:
   $81.3\% \text{ yield} = 0.813 \times 6 \text{ mol CO2} = 4.88 \text{ mol CO2}$

4. Convert moles of CO2 to grams:
   $4.88 \text{ mol} \times \frac{44.01 \text{ g/mol}}{1 \text{ mol}} = 214.8 \text{ g CO2}$

So, 214.8 grams of CO2 would be produced.

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

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