Question

What is the mass of strontium chloride that reacts with 300.0 g of sulfuric acid?

Answer 1

To find the mass of strontium chloride that reacts with 300.0 g of sulfuric acid, you need to use the balanced chemical equation and stoichiometry calculations.

Step-by-step explanation:

To find the mass of strontium chloride that reacts with 300.0 g of sulfuric acid, we need to know the balanced chemical equation for the reaction between strontium chloride (SrCl2) and sulfuric acid (H2SO4). Once we have the balanced equation, we can use stoichiometry to calculate the mass.

The balanced chemical equation for the reaction is:

SrCl2 + H2SO4 → SrSO4 + 2HCl

From the balanced equation, we see that 1 mole of SrCl2 reacts with 1 mole of H2SO4. So, we need to convert the given mass of H2SO4 to moles:

Moles of H2SO4 = Mass of H2SO4 / Molar mass of H2SO4

After finding the moles of H2SO4, we can use the stoichiometric ratio to determine the moles of SrCl2 that react with it. Finally, we can convert the moles of SrCl2 to grams using its molar mass.

Answer 2

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