An aqueous solution of silver nitrate contains 5.00 g of silver nitrate. To this solution is added sufficient barium chloride to precipitate all the silver as silver chloride. What mass, in grams, of dry silver chloride can be obtained?

You can do this one of several ways, first, as a reaction to precipate the silver.

The second, is just to figure out how much silver is in 5g AgNO3. This is much easier. The percent of Ag in AgNO3 is

Percent silver: atomicmassSilver/formula mass AgNO3

To find the mass of silver chloride that can be obtained, we need to first understand the reaction that takes place between silver nitrate (AgNO3) and barium chloride (BaCl2). The reaction can be written as follows:

AgNO3 + BaCl2 -> AgCl + Ba(NO3)2

From the balanced chemical equation, we can see that 1 mole of silver nitrate reacts with 1 mole of barium chloride to produce 1 mole of silver chloride. Furthermore, we can use the molar masses of the compounds involved to calculate the amount of product formed.

1. Calculate the number of moles of silver nitrate:
The molar mass of AgNO3 (silver nitrate) is:
Ag: 107.87 g/mol
N: 14.01 g/mol
O: 16.00 g/mol (x3)

Adding these atomic masses together:
107.87 + 14.01 + (16.00 × 3) = 169.87 g/mol

Given that you have 5.00 g of silver nitrate, you can calculate the number of moles:
Moles = Mass / Molar mass
Moles = 5.00 g / 169.87 g/mol ≈ 0.0294 mol

2. Determine the number of moles of silver chloride formed:
From the balanced equation, we know that 1 mole of silver nitrate produces 1 mole of silver chloride. Therefore, the number of moles of silver chloride formed is also 0.0294 mol.

3. Calculate the mass of silver chloride:
The molar mass of AgCl (silver chloride) is:
Ag: 107.87 g/mol
Cl: 35.45 g/mol

Adding these atomic masses together:
107.87 + 35.45 = 143.32 g/mol

Now, we can calculate the mass of silver chloride formed:
Mass = Moles × Molar mass
Mass = 0.0294 mol × 143.32 g/mol ≈ 4.21 g

Therefore, the mass (in grams) of dry silver chloride that can be obtained is approximately 4.21 g.