Given:
Mass of sodium chloride (NaCl) = 5.85 g
Mass of silver chloride (AgCl) = 14.35 g
Mass of sodium nitrate (NaNO3) = 8.5 g

To find:
Mass of silver nitrate (AgNO3) that will react with NaCl.

Solution:
1. Write the balanced chemical equation for the reaction between NaCl and AgNO3 to form AgCl and NaNO3.
NaCl + AgNO3 → AgCl + NaNO3

2. Find the molar masses of the compounds involved in the reaction.
Molar mass of NaCl = 58.44 g/mol
Molar mass of AgNO3 = 169.87 g/mol
Molar mass of AgCl = 143.32 g/mol
Molar mass of NaNO3 = 84.99 g/mol

3. Convert the given masses of NaCl, AgCl, and NaNO3 into moles using their respective molar masses.
Moles of NaCl = 5.85 g / 58.44 g/mol = 0.1 mol
Moles of AgCl = 14.35 g / 143.32 g/mol = 0.1 mol
Moles of NaNO3 = 8.5 g / 84.99 g/mol = 0.1 mol

4. Use the mole ratio between NaCl and AgNO3 from the balanced chemical equation to find the moles of AgNO3 that reacted with NaCl.
From the balanced equation, 1 mole of NaCl reacts with 1 mole of AgNO3.
Therefore, moles of AgNO3 = moles of NaCl = 0.1 mol.

5. Convert the moles of AgNO3 into mass using its molar mass.
Mass of AgNO3 = moles of AgNO3 x molar mass of AgNO3
Mass of AgNO3 = 0.1 mol x 169.87 g/mol = 17.0 g

Therefore, the mass of silver nitrate that will react with 5.85 g of sodium chloride to produce 14.35 g of silver chloride and 8.5 g of sodium nitrate is 17.0 g. Hence, the correct option is (c) 17.0 g.

Given:

Mass of sodium chloride (NaCl) = 5.85 g

Mass of silver chloride (AgCl) produced = 14.35 g

Mass of sodium nitrate (NaNO3) produced = 8.5 g

To find:

Mass of silver nitrate (AgNO3) needed

Solution:

1. Write the balanced chemical equation for the reaction:

AgNO3 + NaCl → AgCl + NaNO3

2. Calculate the molar mass of NaCl:

Molar mass NaCl = 23 + 35.5 = 58.5 g/mol

3. Calculate the number of moles of NaCl:

Number of moles NaCl = Mass of NaCl / Molar mass NaCl

Number of moles NaCl = 5.85 g / 58.5 g/mol = 0.1 mol

4. Calculate the number of moles of AgCl produced:

From the balanced chemical equation, we can see that 1 mole of NaCl reacts with 1 mole of AgNO3 to produce 1 mole of AgCl.

So, the number of moles of AgCl produced = 0.1 mol

5. Calculate the mass of AgCl produced:

Mass of AgCl = Number of moles of AgCl produced x Molar mass of AgCl

Molar mass AgCl = 107.87 + 35.45 = 143.32 g/mol

Mass of AgCl = 0.1 mol x 143.32 g/mol = 14.32 g

6. Calculate the number of moles of NaNO3 produced:

From the balanced chemical equation, we can see that 1 mole of NaCl reacts with 1 mole of AgNO3 to produce 1 mole of NaNO3.

So, the number of moles of NaNO3 produced = 0.1 mol

7. Calculate the mass of NaNO3 produced:

Mass of NaNO3 = Number of moles of NaNO3 produced x Molar mass of NaNO3

Molar mass NaNO3 = 22.99 + 14.01 + 3x16.00 = 84.99 g/mol

Mass of NaNO3 = 0.1 mol x 84.99 g/mol = 8.5 g

8. Calculate the number of moles of AgNO3 used:

From the balanced chemical equation, we can see that 1 mole of NaCl reacts with 1 mole of AgNO3.

So, the number of moles of AgNO3 used = 0.1 mol

9. Calculate the mass of AgNO3 used:

Mass of AgNO3 = Number of moles of AgNO3 used x Molar mass of AgNO3

Molar mass AgNO3 = 107.87 + 14.01 + 3x16.00 = 169.87 g/mol

Mass of AgNO3 = 0.1 mol x 169.87 g/mol = 16.99 g

10. Verify the law of conservation of mass:

Mass of reactants = Mass of AgNO3 + Mass of NaCl = 16.99 g + 5.85 g = 22.84 g

Mass of products = Mass of Ag