Answer to Question #140121 in General Chemistry for Linh

Question

The toxic gas, hydrogen fluoride, HF, is produced from the double displacement reaction of calcium fluoride, CaF2, and concentrated sulfuric acid H2SO4. Starting with 10.0 g of CaF2 and 15.5 g of H2SO4, how many grams of HF will be produced? Which is limiting reactant? What grams of the excess reactant will be left after the reaction?

Answer 1

CaF₂ + H₂SO₄ → 2HF + CaSO₄

M(CaF₂) = 78.07 g/mol

M(H₂SO₄) = 98.08 g/mol

M(HF) = 20.01 g/mol

n = m/M

n(CaF₂) $= \frac{10.0}{78.07} = 0.128 \;mol$

n(H₂SO₄)$= \frac{15.5}{98.08} = 0.158 \;mol$

Proportion according to the reaction:

1 mol CaF₂ : 1 mol H₂SO₄

Actual amount of reactants:

0.128 mol CaF₂ : 0.158 mol H₂SO₄

So, H₂SO₄ is in excess and CaF₂ is the limiting reactant.

n(HF) = 2n(CaF₂) $= 2 \times 0.128 = 0.256 \;mol$

m(CaF₂) $= 0.256 \times 20.01 = 5.12 \;g$

n(H₂SO₄) = 0.128 mol (needs for reaction)

∆n(H₂SO₄) = 0.158 – 0.128 = 0.03 mol

Mass of excess reactant:

m(H₂SO₄) $= 0.03 \times 98.08 = 2.94 \;g$