how many grams of silver carbonate will precipitate when excess ammonium carbonate soln is added to 41.0 mL of a .518 M silver nitrate solution?

To determine how many grams of silver carbonate will precipitate when excess ammonium carbonate solution is added, we first need to write and balance the chemical equation for the reaction between silver nitrate and ammonium carbonate.

The balanced chemical equation is as follows:
2 AgNO3 + (NH4)2CO3 → Ag2CO3 ↓ + 2 NH4NO3

From the balanced equation, we can see that the stoichiometric ratio between silver nitrate (AgNO3) and silver carbonate (Ag2CO3) is 2:1. This means that for every 2 moles of AgNO3, we will get 1 mole of Ag2CO3.

Now, let's calculate the moles of AgNO3 present in the given solution:

Moles of AgNO3 = Concentration (M) × Volume (L)
= 0.518 mol/L × 0.0410 L
≈ 0.021238 mol

Since the stoichiometric ratio is 2:1, we know that we will get half as many moles of Ag2CO3 as AgNO3.

Moles of Ag2CO3 = Moles of AgNO3 ÷ 2
= 0.021238 mol ÷ 2
≈ 0.010619 mol

Now, we can calculate the mass of Ag2CO3 using its molar mass.

Molar mass of Ag2CO3 = (2 × Atomic mass of Ag) + Atomic mass of C + (3 × Atomic mass of O)
= (2 × 107.87 g/mol) + 12.01 g/mol + (3 × 16.00 g/mol)
= 275.74 g/mol

Mass of Ag2CO3 = Moles of Ag2CO3 × Molar mass of Ag2CO3
= 0.010619 mol × 275.74 g/mol
≈ 2.9379 g

Therefore, approximately 2.9379 grams of silver carbonate will precipitate when excess ammonium carbonate solution is added to 41.0 mL of a 0.518 M silver nitrate solution.