Chemistry - Stoichiometry Question with Solution | TestHub

The reaction is BaCl₂(aq) + H₂SO₄(aq) -> BaSO₄(s) + 2HCl(aq).

Initial moles: BaCl₂ = $0.1\times 0.4=0.04$ mol; H₂SO₄ = $0.1\times 0.6=0.06$ mol.

BaCl₂ is the limiting reactant.

After reaction:

Ba²⁺: 0 mol (precipitated as BaSO₄)

SO₄²⁻: $0.06-0.04=0.02$ mol (from excess H₂SO₄)

H⁺: $2\times 0.04$ mol (from HCl) + $2\times 0.02$ mol (from excess H₂SO₄) = $0.08+0.04=0.12$ mol.

Cl⁻: $2\times 0.04=0.08$ mol.

Total volume = $0.4+0.6=1.0$ L.

Final molarities:

$[B{a}^{2+}]=0$ M.

$[S{O}_4^{2-}]=0.02/1.0=0.02$ M.

$[H^{+}]=0.12/1.0=0.12$ M.

$[C{l}^{-}]=0.08/1.0=0.08$ M.

Option D is correct.