How many grams of iron(II)sulfide is required to make 225 g of sulfuric acid,H₂SO₄? The reactions involved are given below. 4 FeS + 7O₂ → 2 Fe₂O₃ + 4 SO₂ 2 SO₂ + O₂ → 2 SO₃ SO₃ + H₂O → H₂SO₄ A)101 g B)251 g C)126 g D)202 g E)502 g

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First, we need to understand that the final product is sulfuric acid (H₂SO₄). The given reactions show the process of converting iron(II) sulfide (FeS) into sulfuric acid through multiple steps. The molar mass of H₂SO₄ is approximately 98 g/mol, meaning 225 g of H₂SO₄ corresponds to approximately 2.3 moles (225 g / 98 g/mol). The stoichiometry of the first reaction shows that 4 moles of FeS produce 4 moles of SO₂, which eventually leads to the production of 2 moles of H₂SO₄. Therefore, to produce 2.3 moles of H₂SO₄, we need 2.3 moles of FeS, considering the stoichiometry of the entire process. The molar mass of FeS is approximately 88 g/mol, so the required mass of FeS is 2.3 moles * 88 g/mol = 202.4 g, which rounds to 202 g.

How Many Grams of Iron(II)sulfide Is Required to Make 225