Question

would it be:

O.2576/0.7086=0.3635
0.3635 X 87.65= actual yield?


A 2.30-mL sample of methane gas (density 0.112 g/cm3) was completely combusted. If the percent yield of each of the combustion products is 87.65%, calculate the actual yield in grams of carbon dioxide gas.

* chemistry - DrBob222,

CH4 + 2O2 ==> CO2 + 2H2O
g CH4 = 2.30 mL x 0.112 g/mL = 0.2576
moles CH4 = 0.2576/16 = 0.0161

moles CO2 must be 0.0161
g CO2 = moles x molar mass = theoretical yield.
Then (actual g/theoretical)*100 = %yield
You know theoretical and %yield, solve for actual yield.

Answer 1

<b>No. Are you trying to say that you start with 0.2576g CH4 and you end up with 0.3635 x 87.65 = 31.86 g CO2? The theoretical yield is ONLY 0.708 grams; how can the actual yield exceed that by 44 times.</b>

CH4 + 2O2 ==> CO2 + 2H2O
g CH4 = 2.30 mL x 0.112 g/mL = 0.2576
moles CH4 = 0.2576/16 = 0.0161

moles CO2 must be 0.0161<b> also from the coefficients in the balanced equation.</b>
g CO2 = moles x molar mass = theoretical yield.
<b>0.0161 moles CO2 x (44 g CO2/mole CO2) = 0.708 g CO2. This is the theoretical yield. </b>
Then (actual g/theoretical)*100 = %yield
You know theoretical and %yield, solve for actual yield.
<b>
(actual yield/theor yield)*100 = 87.65%
(actual yield/0.708 g)*100 = 87.65%
(actual yield/0.708) = 0.8765
(actual yield) = 0.8765*0.708 = 0.6209 which rounds to 0.621 to three s.f.
</b>