Answer to Question #240034 in General Chemistry for Genesis Ny

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:

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