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If 39.1 g of AgNO₃ react with 28.6 g of H₂SO₄ according to the unbalanced equation below, what is the mass in grams of Ag₂SO₄ that could be formed?

AgNO₃(aq) + H₂SO₄(aq) → Ag₂SO₄(s) + HNO₃(aq)

Answer :

Mass of Ag₂SO₄ is 71.77 g .The mole ratio between AgNO₃ and Ag₂SO₄, The limiting reactant, and convert moles of Ag₂SO₄ to grams.

The mass of Ag₂SO₄ that could be formed, we need to find the limiting reactant. First, calculate the moles of AgNO₃ and H₂SO₄ using their respective molar masses. Then, determine the mole ratio between AgNO₃ and Ag₂SO₄ in the balanced equation. From there, find the moles of Ag₂SO₄ that could be formed, and finally convert it to grams.

Given: AgNO₃ = 39.1 g, H₂SO₄ = 28.6 g

Molar mass of AgNO₃ = 169.87 g/mol

Molar mass of H₂SO₄ = 98.09 g/mol

Mole ratio between AgNO₃ and Ag₂SO₄ in the balanced equation is 1:1

Moles of AgNO₃ = 39.1 g / 169.87 g/mol = 0.230 moles

Moles of H₂SO₄ = 28.6 g / 98.09 g/mol = 0.292 moles

Since the mole ratio is 1:1, the limiting reactant is AgNO₃. Therefore, 0.230 moles of Ag₂SO₄ = 0.230 moles x (molar mass of Ag₂SO₄ = 311.83 g/mol

Learn more about limiting reactant here:

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