Question

If 67.2 g of AgNO₃ react with 28.6 g of H₂SO₄ according to this unbalanced equation below, what is the mass in grams of Ag₂SO₄ that could be formed?

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

A) 64.3 g
B) 69.1 g
C) 74.8 g
D) 79.6 g

Answer 1

The mass of Ag2SO4 that can be formed when 67.2 g of AgNO3 react with 28.6 g of H2SO4 is 45.70 g, based on the stoichiometry of the balanced chemical equation and the molar masses of the reactants and products.

Step-by-step explanation:

The student is asking what mass of Ag2SO4 can be formed when 67.2 g of AgNO3 reacts with 28.6 g of H2SO4. To solve this, we must first balance the chemical equation: 2 AgNO3 (aq) + H2SO4 (aq) → Ag2SO4 (s) + 2 HNO3 (aq). Then, we'll calculate the moles of each reactant. The molar mass of AgNO3 is 169.87 g/mol and that of H2SO4 is 98.08 g/mol. This means we have 0.395 mol AgNO3 and 0.291 mol H2SO4. From the stoichiometry of the balanced equation, H2SO4 is the limiting reactant, as we need only 0.291/0.395 ≈ 0.1465 mol of Ag2SO4 (half of the moles of AgNO3). The molar mass of Ag2SO4 is 311.8 g/mol, so multiplying the moles of Ag2SO4 by this molar mass gives us the mass of Ag2SO4 that can be formed: 0.1465 mol × 311.8 g/mol = 45.70 g.