Use the reaction: 2AgNO₃(aq) + H₂SO₄(aq) → Ag₂SO₄(s) + 2H₂O(l). What volume of 0.123 M AgNO₃(aq) is needed to form 0.657 g of Ag₂SO₄(s)? A)34.2 L B)17.1 mL C)34.3 mL D)10.7 mL E)53.4 mL

Question

Quizplus · Accepted Answer

From the balanced chemical equation, we can see that 2 moles of AgNO₃ react with 1 mole of Ag₂SO₄ to produce a 1:1 ratio of product to reactant. Therefore, we can use stoichiometry to find the moles of AgNO₃ needed to produce 0.657 g of Ag₂SO₄: moles of Ag₂SO₄ = mass / molar mass = 0.657 g / (107.87 g/mol + 32.06 g/mol + 4(16.00 g/mol)) = 0.00454 mol Since the mole ratio of AgNO₃ to Ag₂SO₄ is 2:1, we need half as many moles of AgNO₃: moles of AgNO₃ = 0.00454 mol / 2 = 0.00227 mol Now we can use the molarity of AgNO₃ to calculate the volume needed: volume of AgNO₃ = moles / molarity = 0.00227 mol / 0.123 mol/L = 0.0343 L = 34.3 mL (rounded to three significant figures) Therefore, the correct answer is C) 34.3 mL.

Use the Reaction: 2AgNO₃(aq) + H2SO₄(aq) → Ag2SO₄(s) + 2H2O(l)

Use the Reaction: 2AgNO₃(aq) + H₂SO₄(aq) → Ag₂SO₄(s) + 2H₂O(l)