Question

Hydrogen fluoride, HF (g), is a highly toxic gas. It is produced according to the following balanced chemical equation:

CaF₂( s)+H₂SO₄(aq)→2HF(g)+CaSO₄( s)
Determine the limiting reactant when 1.00 g of calcium fluoride, CaF₂( s), reacts with 15.5 g of sulfuric acid, H₂SO₄(aq).

Answer 1

The limiting reactant when reacting 1.00 g of calcium fluoride with 15.5 g of sulfuric acid to produce HF gas is calcium fluoride, as it is present in lesser molar quantity than required by H2SO4 based on the balanced equation.

Step-by-step explanation:

The student is asking how to determine the limiting reactant when reacting calcium fluoride with sulfuric acid to produce hydrogen fluoride gas. To find the limiting reactant, we calculate moles of each reactant based on the balanced chemical equation. The molar mass of CaF2 is approximately 78.07 g/mol, and for H2SO4 it is approximately 98.079 g/mol. With the given masses, we have 1.00 g CaF2 (0.0128 mol) and 15.5 g H2SO4 (0.158 mol).

According to the equation, the molar ratio of CaF2 to H2SO4 is 1:1. Therefore, 0.0128 mol of CaF2 would require 0.0128 mol of H2SO4, but we have 0.158 mol of H2SO4, which is more than sufficient. Thus, calcium fluoride is the limiting reactant.