A asked in Science & MathematicsChemistry · 3 years ago

Chemistry percent yield problem involving STP?

Consider the reaction:

2 SO2(g) + O2(g) -> 2 SO3(s)

A.) If 285.5 mL of SO2 is allowed to react with 158.9 ml of O2 (both measured at STP), what is the limiting reactant and the theoretical yield of SO3?

I got the correct answers given from the book for this part, which are that SO2 is the limiting reactant and the theoretical yield is 0.0127 mol of SO3.

B.) If 2.805 g of SO3 is collected (measured at STP), what is the percent yield for the reaction?

I need some help with this part. The answer given in 65.6 percent. I know how to calculate percent yield, but I don't know how to get that answer for this question. Could someone answer and explain please?

Thanks

1 Answer

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  • Dr W
    Lv 7
    3 years ago
    Favorite Answer

    I cover stoichiometry here..

    https://answers.yahoo.com/question/index?qid=20151...

    read through it if you don't understand what I'm doing

    **********

    your balanced equation

    .. 2 SO2(g) + 1 O2(g) -> 2 SO3(s)

    IF SO2 was the limiting reagent, we would form this much SO3

    .. 0.2855L SO2.... ... 1 mol SO2.. .. .. .. 2 mol SO3.. .. 80.07g SO3

    --- ---- ----- ---- x ----- ---- ---- ---- ----- x ----- ----- ---- x ---- ----- ---- --- = 1.0201g SO3.. (4 sig figs)

    .. ... .. ..1. ...... ... 22.41L SO2.@STP.. 2 mol SO2.. .... 1 mol SO3

    IF O2 was the limiting reagent, we would form this much SO3

    .. 0.0.1589L O2.... ... 1 mol O2.. .. ... .. 2 mol SO3.. .. 80.07g SO3

    --- ---- ----- ---- x ----- ---- ---- ---- ----- x ----- ----- ---- x ---- ----- ---- --- = 1.1355g SO3.. (4 sig figs)

    .. ... .. ..1. ...... ... 22.41L O2.@STP..... 1 mol O2.. .... 1 mol SO3

    since the SO2 gave less SO3, the SO2 is the limiting reagent and the theoretical yield of SO3 = 1.020g

    in terms of moles.. theoretical yield = 0.2855 / 22.41 *2 / 2 = 0.0127 mol SO3

    so we're in agreement on part (a) right?

    ***********

    part (b) has a problem.

    .. (1) SO3 is indeed a solid at STP. it's mp = 17°C @ 1atm of pressure (STP = 0°C + 1atm)

    .. (2) why is "measured at STP" included in the "if 2.805g of SO3 is collected" if we all know

    .. .. ..that SO3(s) was formed?.. yup.. that's weak.. could have been there to throw you off

    .. .. ..or to verify that SO3 was a solid, but it's still a tiny bit unusual

    .. (3) the theoretical yield = 0.0127 mol SO3 according to your book

    ... .. .for 2.805g to be 65.6% of the theoretical yield... either

    .. .. . ..(a) the molar mass used to convert 0.0127 was 335.6 g/mol

    .. .. .. .(b) the book has a typo

    .. .. . . (c) maybe the units on 2.805 aren't "g" but something else.

    .. .. . ..(d) maybe your looking at the wrong "given answer".. problem 133 instead of 131 for example.

    let's do this.. let's assume 65.6% is correct

    .. % yield = actual yield / theoretical yield * 100%

    .. .actual yield = % yield / 100% * theoretical yield

    .. .actual yield = 65.6% / 100% * 1.020g = 0.6619g actual yield

    hmm...

    .. 2.805 isn't just a single digit typo.... nor is it a decimal point being off.

    .. furthermore, the theoretical yield = 1.020g... not possible to form 2.805g

    .. the molar mass probably isn't the issue... hard to go from this

    .. ... 1 * 32.07 + 3 * 16 = 80.07

    . .to this

    ... .. .___ * ___ + ___ * ___ = 335.6

    .. so I doubt that's the case

    are you sure you're looking at the right numbers?

    If so, let' your teacher know

    .. % yield = 0.6619g / 1.020g * 100% = 65.6%

    ..

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