The products of complete combustion of ethanol, C2H5OH, are shown in the unbalanced equation:

Question

C2H5OH (l) + O2 (g) ----> CO2 (g) + H2O (l) 

 

This reaction has a CO2 percent yield of 62.3%

 

What will be the mass of CO2 EXPERIMENTALLY obtained when 44.5g O2 and 24.6g C2H5OH are reacted? 

 

Molar masses: C2H5OH = 46.07 g/mol; O2= 32.00 g/mol; CO2= 44.01 g/mol; H2O= 18.016 g/mol

 

 

Answer 1

anytime you have a problem involving how much of this or that is consumed, produced, needed.  OR  what is the % yield or actual yield.  You have yourself a problem in reaction "stoichiometry".

 

Here are the 6 steps to solve stoichiometry problems

.. (1) write a balanced equation

.. (2) convert everything given to MOLES

.. (3) determine limiting reagent

.. (4) convert moles limiting reagent to moles of other chemicals via

.. .. ..coefficients of balanced equation.

.. (5) convert moles back to mass.  This is theoretical mass.  AKA

... .. .theoretical yield

.. (6) % yield = (actual mass recovered / theoretical mass) x 100%

 

the basic idea being the coefficients of a balanced equation are in mole ratios.  So if we know what those ratios are, we can use them to convert between moles of the different chemicals.  The maximum mass that can be produced is the theoretical mass.  The actual mass recovered takes losses and inefficiencies and incomplete reactions in account.

 

so here we go.  Step by step

 

*** 1 ***

1 C2H5OH + 3 O2 --> 2 CO2 + 3 H2O

 

*** 2 ***

mass O2 = 44.5g O2 x (1 mol O2 / 32.00g O2) = 1.391 mol O2 (with 1 extra sig fig this is an intermed. step)

 

mass C2H5OH = 24.6g x (1 mol / 46.07g) = 0.5340.. (same deal.  I'm carrying 1 extra sig fig)

 

*** 3 ***

from the balanced equation, 1 mol ethanol (EtOH.. that C2H5OH you see) reacts with 3 moles O2.  So we NEED this much O2 to react with all the alcohol.

.. 0.5340mol EtOH x (3 mol O2 / 1 mol EtOH) = 1.602 mol O2

 

since we NEED 1.602 mol O2 and we only HAVE 1.391 mol O2 available, we don't have enough O2 to completely consume all the C2H5OH and therefore O2 is the limiting reagent.

 

note... this is unusual.  The reason is if we don't have enough O2, we tend to form other chemicals like CO instead of CO2.  So most problems have XS (eXcesS) O2 and the fuel (the alcohol in this case) is the L.R... In the event that you encounter a problem where O2 is the L.R... you should immediately recheck your steps 1-3.  If everything still works out (like it does here) you might want to take note that you're assuming the only products are CO2 and H2O.

 

continuing

 

*** 4 ***

here we're converting moles of the limiting reagent to moles of the final products (CO2 is the product we're interested in in this case).  from the balanced equation... 3 O2 --> 2 CO2

 

so that theoretical moles CO2 = 1.391 mol O2 x (2 mol CO2 / 3 mole O2) = 0.9273

 

*** 5 ***

theoretical mass.

.. th.mass CO2 = 0.9273 mol CO2 x (44.01g CO2 / mol CO2) = 40.81g CO2

 

*** 6 ***

and finally

.. % yield = (actual / theoretical) x 100%

 

rearranging

.. actual = (% yield / 100%) x theoretical = (62.3% / 100%) x 40.81g CO2 = 25.4g CO2... (3 sig figs)

 

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now.. if you notice, there's a whole lot of calculating then carrying numbers over to the next line.  We can streamline this solution by using "dimensional analysis" and this concept.  We convert all the reactants to a given product. Whichever yields the least amount of product must be the limiting reagent.

 

like this

 

IF O2 is the L.R.

 

    44.5g O2       1 mol O2      2 mol CO2     44.01g CO2

  ------- ------- x ------- ------- x -------- ------- x -------- --------- = 40.80g CO2

          1             32.00g O2     3 mol O2      1 mol CO2

 

IF C2H5OH is the L.R.

 

    24.6g EtOHO2    1 mol EtOH       2 mol CO2      44.01g CO2
   ---------- --------- x --------- -------- x -------- -------- x -------- -------- = 47.00g CO2
              1                46.07g EtOH    1 mol EtOH     1 mol CO2

 

 

since O2 gave less CO2, O2 must be the limiting reagent

 

and btw.. you enter those calcs in your calculator like this

   44.5 / 32 x 2 / 3 x 44.01 =

and

   24.6 / 46.07 x 2 x 44.01 = 

all linear just like I show. then round to 3 sig figs.  Can you see why?

 

 

one last thing.  Take a look at the steps.  I start with my mass on the left.

.. (a) then I multiply by 1 / molar mass to convert to moles

.. (b) then I multiply by the coefficients of the balanced equation to convert to moles

.. .. ..of other chemicals

.. (c) then I multiply by molar mass to convert back to mass

steps (2), (4) and (5) above

then use the results to solve step (3) above.

 

see how that fits together?  

 

finishing the problem is the same

.. actual mass recovered = (62.3 / 100) x 40.80g = 25.4g CO2.. (3 sig figs)

 

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either way, the answer is 25.4g.

 

My recommendation is 

.. (a) you practice solving the problem with the step by step approach until you know those steps cold

.. (b) you then abandon that approach and solve the problems via dimensional analysis from here on.

 

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good luck

 

and PS.. if you don't know dimensional analysis... you need to learn it.  it's the most important tool you'll ever use in general chemistry.

 

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you're very welcome Lara.