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Limiting Reactants

  1. Jul 27, 2015 #1
    The problem statement, all variables and given/known data

    I wanted to check my work on this to make sure I'm understanding limiting reactants.

    You added 10.0g CaF2 to 10.0g H2SO4 to 1L H2O
    1. How much CaSO4 can you possibly make?
    2. Which is the limiting reagent?
    3. Which reactant is in excess? By how much?

    The attempt at a solution

    CaF2 + H2SO4 -> CaSO4 + 2HF

    10g CaF2(1 mol CaF2/78g CaF2) = 0.13 mol CaF2
    10g H2SO4(1mol H2SO4/98H2SO4) = 0.10 mol H2SO4

    I believe this answers questions 2 and 3. The limiting reactant is H2SO4, meaning the reactant that is in excess is CaF2. For this part since it is 1:1 I assumed that I'd subtract 0.10 from 0.13 and I was left with 0.03 mol CaF2. Converting that back to grams:

    0.03 mol CaF2(78g CaF2/1 mole CaF2) = 2.34g CaF2

    Then for 1 I got:
    10g H2SO4(1 mole H2SO4/98g H2SO4)(1 mole CaSO4/1 mole H2SO4)(136g CaSO4/1 mole CaSO4) = 14g CaSO4

    1. 14g CaSO4
    2. H2SO4
    3. 2.34g CaF2

    I don't mind if my math is a tad off, I'm more concerned with the concept at this point. Thanks for your time, again. :)
     
  2. jcsd
  3. Jul 27, 2015 #2
    It looks OK to me.

    Chet
     
  4. Jul 27, 2015 #3
    Thanks Chet, appreciate you. :)
     
  5. Jul 28, 2015 #4

    Borek

    User Avatar

    Staff: Mentor

    Logic is OK, math not so.

    Watch your significant figures. 14g of CaSO4 is in the right ballpark, but it should have 3 sigfigs, not 2.

    Don't use rounded down amounts in your calculations. Use as many digits as you have (but report the rounded down values). Because of rounding errors your mass of excess CaF2 is off by about 15%.
     
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