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Question

If 42.3 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)

Answer 1

First, we must balance the equation:

2AgNO₃(aq) + H₂SO₄(aq) ==> Ag₂(SO₄)(s) + 2HNO₃(aq) .. balanced equation

Next, we must find the limiting reactant since we are given amounts of both reactants. One way to do this is to divided the moles of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less represents the limiting reactant.

For AgNO₃: 42.3 g AgNO₃ x 1 mol / 170. g = 0.2488 mols (÷2->0.12)

For H₂SO₄: 28.6 g H₂SO₄ x 1 mol / 98.1 g = 0.2915 mols (÷1->0.29)

Since 0.12 is less than 0.29, AgNO₃ is the limiting reactant. We will use moles AgNO₃(0.2488 mols) for the rest of the calculation.

Using the mols of AgNO₃ and the stoichiometry of the balanced equation, we will find moles of Ag₂SO₄ formed. Then using molar mass of Ag₂SO₄ (311.8 g / mol), we'll convert moles to grams for our answer.

0.2488 mol AgNO₃: x 1 mol Ag₂SO₄ / 2 molAgNO₃: = 0.1244 mols Ag₂SO₄

0.1244 mols Ag₂SO₄ x 311.8 g / mol = 38.8 g Ag₂SO₄ (3 sig. figs.)