Limiting Reactant & Excess Reactant Calculation for CaF2 + H2SO4

[in the rye]

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. :)

 

[Chestermiller]

It looks OK to me.

Chet

 

[in the rye]

It looks OK to me.

Chet

Thanks Chet, appreciate you. :)

 

[Borek]

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%.