Limiting Reactants

  • Thread starter in the rye
  • Start date
  • #1
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Homework Statement

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. :)
 

Answers and Replies

  • #4
Borek
Mentor
28,781
3,259
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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