Help with chemistry/how to calculate percent yield?

So the question is: Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, the reaction of 5.0 grams of O2 and 6.0 grams of S can theoretically produce how many grams of SO3?

I balanced the equation 2S+3O2 -> 2SO3

I also have the molar masses for the reactants but I m not sure how to set up the rest of the problem. I m not looking for someone to just give me the answer, but an explanation and steps would be really helpful! Thanks!

1 Answer

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  • Dr W
    Lv 7
    5 years ago
    Favorite Answer

    anytime you get a problem in "stoichiometry".. i.e.. how much of this or that is reacted or produced.. or.. what is the yield or % yield... follow these 6 steps. MEMORIZE THEM.

    steps to solving stoichiometry problems

    .. (1) write a balanced equation

    .. (2) convert everything given to moles

    .. (3) determine limiting reagent

    .. (4) convert moles limiting reagent to moles other species

    .. (5) convert moles to mass. This is theoretical mass.. aka theoretical yield

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

    the idea being, the coefficients of the balanced equation are in MOLE ratios. We determine those coefficients first, then use the ratios to convert between different chemicals.

    now.. there's the long way of doing this.. following each step 1 step at a time. And the shortcut via dimensional analysis. I'll show you both. I strongly encourage you to learn dimensional analysis (factor label method) pronto and use it if you haven't already done so. it will make your chemistry life much easier.

    the long way

    *** 1 ***

    balanced equation

    .. 2 S + 3 O2 ---> 2 SO3.... that is correctly balanced.. excellent so far

    *** 2 ***

    conversion to moles

    .. 5.0g O2 x (1 mol O2 / 32.0g O2) = 0.156mol O2

    .. 6.0g S x (1 mol S / 32.06g S) = 0.187mol S

    notice I'm carrying along 1 extra sig fig? this is an intermediate step, the rules of sig figs allow carrying an extra sig fig during intermediate steps. I recommend you do this.

    *** 3 ***

    there are 3 common ways to determine limiting reagent

    .. (1) chose a reactant. Determine moles of the OTHER reactant required to

    .. .. ..completely consume the first reactant. If you NEED more of the other reactant

    .. .. . than you have available, the OTHER reactant is limiting. If you HAVE more

    .. . ...of the OTHER reactant than you need, the first reactant (the one you chose)

    .. .. ..is the limiting reagent

    .. (2) divide moles available by coefficients of the balanced equation for

    .. .. ..all reactants. The smallest ratio belongs to the limiting reagent

    .. (3) via dimensional analysis, convert both reactants to products. whichever

    .. .. ..reactant gives the least amount of product must be the limiting reagent

    I'll show the first two methods now, and save the 3rd for the dimensional analysis method at the end. Here we go

    method (1)

    .. I choose, S

    .. from the balanced equation , 2 mol S reacts with 3 mol O2

    .. so that moles O2 NEEDED is

    .. .. 0.187mol S x (3 mol O2 / 2 mol S) = 0.280mol O2

    .. from part(2) moles O2 AVAILABLE = 0.156

    .... .we NEED more O2 to react all the S than we HAVE so O2 is limiting

    method (2)

    .. mol S / coefficient S = 0.187 / 2 = 0.0935

    .. mol O2 / coefficient O2 = 0.156 / 3 = 0.052

    since the mol O2 / coefficient O2 ratio < mole S / coefficient S ratio

    O2 is the limiting reagent

    *** 4 ***

    from the balanced equation, 3 mol O2 ---> 2 mol SO3 so that

    .. 0.156mol O2 x (2 mol SO2 / 3 mol O2) = 0.104 mol SO3

    *** 5 ***

    theoretical mass

    .. 0.104mol SO3 x (80.06g SO3 / mol SO3) = 8.33g SO3

    *** 6 ***

    we skip this.. we only do this step if we're given an actual mass collected (or asked to calculate one).

    *********

    and the final answer is

    .. theoretical mass SO3 = 8.3g... (2 sig figs)

    **********

    **********

    **********

    **********

    via dimensional analysis..

    we convert BOTH reactants to products... whichever gives less amount of product is the limiting reagent.

    IF S is the limiting reagent...

    .. .6.0g S.. .1mol S.. 2 mol SO3.. 80.06g SO3

    .. --- ----- x --- ----- x -- ----- ----- x --- ----- ----- = 15g SO3

    .. .. ..1.. ....32.0gS.. 2 mol S.. ... .. 1mol SO3

    IF O2 is the limiting reagent...

    .. .5.0gO2.. .1mol O2... 2 mol SO3.. 80.06g SO3

    .. ----- ----- x ------ ----- x -- ----- ----- x --- ----- ----- = 8.3g SO3

    .. .. ..1.. .... .32.0gO2.... 3 mol O2.. ... 1mol SO3

    since O2 gave less SO3

    O2 is the limiting reagent and 8.3g SO3 is produced

    *********

    now let's break one of those equations down and see how it compares to the 6 steps above

    .. ... .. .. .. .step (2).. ... .... ... ... ... .. step (4)

    ... .. .. . converts to moles.. .. ..converts moles to mass

    .. .. .. ... .. .. .. ↓↓↓↓.. ... ... ... ... ... ... ... ↓↓↓↓↓

    .. .5.0gO2.. .1mol O2... 2 mol SO3.. 80.06g SO3

    .. ----- ----- x ------ ----- x -- ----- ----- x --- ----- ----- = 8.3g SO3

    .. .. ..1.. .... .32.0gO2.... 3 mol O2.. ... 1mol SO3

    .. .. .... ... .... .... .... ... ... .. .↑↑↑↑↑

    .. ... ... ... ... ..... ... .. .. .. ..step (3)

    .. ... .. ... uses coefficients of balanced equation to convert

    .. .. .. .. ... .. between moles of different chemicals

    and... we calculate those 15.0 and 8.3 values like this

    .. 6 / 32 x 80.06 =

    and

    .. 5 / 32 x 2 / 3 x 80.06 =

    all at once in your calculator.

    *********

    you pick which method you like. I recommend doing a couple of problems the long long long way to make sure you understand the concepts. Then use the DA method from here on out. The reason?

    .. (1) precision

    .. (2) built in error check

    .. .. .. .if units cancel properly (follow the canceling along)

    .. .. .. .if unit factors are correct (check backwards from right to left)

    .. .. .. .the problem MUST be setup correctly

    .. (3) better precision.. no intermediate rounding

    .. (4) logical problem setup

    .. (5) easy to demonstrate to your instructor that you know what you're doing

    .. (6) and most importantly.. IT'S FAST.

    in industry, we use that DA method because it works well for multicomponent systems..

    example.. imagine this unbalanced reaction.

    .. H2 + O2 + SiCl4 + SiHCl3 + SiCl3CH3 --> SiO2 + CO2 + HCl

    Imagine the ratio of SiHCl3 to SiCl3CH3 is 60:40

    Imagine the ratio of SiCl4 to (SiHCl3 + SiCl3CH3) is 2:1

    what is the theoretical yield of SiO2 if the feed into the reactor is

    .. 1.00 lbmol / hr SiCl3CH3

    .. 12.5 lbmol / hr H2

    .. 13.0 lbmol / hr O2

    hint..

    .. 2 H2 + 1 O2 ---> 2 H2O

    then

    .. 2 H2O + 1 SiCl4 ---> 1 SiO2 + 4 HCl

    etc

    same process.. write a balanced equation

    convert all reactants to products.. whichever gives the least is the limiting reagent

    and that determines the theoretical yield.

    (btw.. the answer is 27.2kg/hr SiO2)

    ******

    questions about any of this?

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