Final answer:
To determine the percent yield of silver(I) sulfate, we need to calculate the theoretical yield and the actual yield. The theoretical yield is calculated using the moles of the limiting reactant and the stoichiometric ratio. The actual yield is the mass of the dried silver(I) sulfure precipitate. The percent yield is then calculated by dividing the actual yield by the theoretical yield and multiplying by 100.
Step-by-step explanation:
To determine the percent yield of silver(I) sulfate, we need to calculate the theoretical yield and the actual yield.
1. Calculating the theoretical yield:
First, we need to find the moles of AgNO₃ and Li₂SO₄ in the solutions:
Moles of AgNO₃ = volume (L) x concentration (M) = 0.1 L x 0.3 M = 0.03 mol
Moles of Li₂SO₄ = volume (L) x concentration (M) = 0.04 L x 0.1 M = 0.004 mol
Next, we find the limiting reactant by comparing the moles of AgNO3 and Li₂SO₄. The ratio between them is 1:2, so the Li₂SO₄ is the limiting reactant.
Using the stoichiometric ratio from the balanced equation, we can calculate the moles of Ag₂SO₄ formed:
Moles of Ag₂SO₄ = 0.004 mol Li₂SO₄ x (1 mol Ag₂SO₄ / 1 mol Li₂SO₄) = 0.004 mol
Finally, we convert the moles of Ag₂SO₄ to grams using the molar mass:
Mass of Ag₂SO₄ = moles x molar mass = 0.004 mol x 312 g/mol = 1.248g
2. Calculating the percent yield:
The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100:
Percent yield = (actual yield / theoretical yield) x 100
Given that the mass of the dried silver(I) sulfure precipitate is 1.45 g, this is the actual yield. Therefore, the percent yield is:
Percent yield = (1.45 g / 1.248 g) x 100 = 116%