How many grams of sulfuric acid are needed to produce 57.0 grams of water? show all steps of your calculation as well as the final answer. 2naoh+h2so4 > 2na2so4+2h2o

155.16 g.

Explanation:

Firstly, It is considered as a stichiometry problem.From the balanced equation: 2NaOH + H₂SO₄ → 2Na₂SO₄ + 2H₂OIt is clear that the stichiometry shows that 2.0 moles of NaOH reacts with 1.0 mole of H₂SO₄ to give 2.0 moles of Na₂SO₄ and 2.0 moles of H₂O.We must convert the grams of water (57.0 g) to moles (n = mass/molar mass).n = (57.0 g) / (18.0 g/mole) = 3.1666 moles.Now, we can get the number of moles of H₂SO₄ that is needed to produce 3.1666 moles of water.Using cross multiplication:1.0 mole of H₂SO₄ → 2.0 moles of H₂O, from the stichiometry of the balanced equation.??? moles of H₂SO₄ → 3.1666 moles of H₂O.The number of moles of H₂SO₄ that will produce 3.1666 moles of H₂O (57.0 g) is (1.0 x 3.1666 / 2.0) = 1.5833 moles.Finally, we should convert the number of moles of H₂SO₄ into grams (n = mass/molar mass).Molar mass of H₂SO₄ = 98.0 g/mole.mass = n x molar mass = (1.5833 x 98.0) = 155.16 g.

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