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?

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:

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