- #1

in the rye

- 83

- 6

**Homework Statement**

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

You added 10.0g CaF

_{2}to 10.0g H

_{2}SO

_{4}to 1L H

_{2}O

1. How much CaSO

_{4}can you possibly make?

2. Which is the limiting reagent?

3. Which reactant is in excess? By how much?

The attempt at a solution

The attempt at a solution

CaF

_{2}+ H

_{2}SO

_{4}-> CaSO

_{4}+ 2HF

10g CaF

_{2}(1 mol CaF

_{2}/78g CaF

_{2}) = 0.13 mol CaF

_{2}

10g H

_{2}SO

_{4}(1mol H

_{2}SO

_{4}/98H

_{2}SO

_{4}) = 0.10 mol H

_{2}SO

_{4}

I believe this answers questions 2 and 3. The limiting reactant is H

_{2}SO

_{4}, meaning the reactant that is in excess is CaF

_{2}. 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 CaF

_{2}. Converting that back to grams:

0.03 mol CaF

_{2}(78g CaF

_{2}/1 mole CaF

_{2}) = 2.34g CaF

_{2}

Then for 1 I got:

10g H

_{2}SO

_{4}(1 mole H

_{2}SO

_{4}/98g H

_{2}SO

_{4})(1 mole CaSO

_{4}/1 mole H

_{2}SO

_{4})(136g CaSO

_{4}/1 mole CaSO

_{4}) = 14g CaSO

_{4}

1. 14g CaSO

_{4}

2. H

_{2}SO

_{4}

3. 2.34g CaF

_{2}

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. :)