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May 27, 2015

Homework Help: chemistry

Posted by gary on Saturday, October 23, 2010 at 5:16pm.

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.

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