SrCl₂ reacts with conc. H₂SO₄ and gives SrSO₄ and 2HCl as the products. The balanced equation is
SrCl₂ + H₂SO₄ → SrSO₄ + 2HCl
moles = mass / molar mass
molar mass of H₂SO₄ = 98 g mol⁻¹
moles of H₂SO₄ = 300.0 g / 98 g mol⁻¹ = 3.06 mol
Stoichiometric ratio between SrCl₂ and H₂SO₄ is 1 : 1
Hence, moles of SrCl₂ = moles of H₂SO₄
= 3.06 mol
molar mass of SrCl₂ = 158.53 g mol⁻¹
Hence, mass of SrCl₂ = 3.06 mol x 158.53 g mol⁻¹
= 485.1 g
Hence, the mass of SrCl₂ needed to react with H₂SO₄ is 485.1 g