chlorine is precipitated as silver chloride by reaction of silver nitrate with sodium chloride in an acid solution .how much sodium chloride in grams is required to produce 0.25g of silver chloride

a)0.102g
b)0.613
c)1.019
d)1.631g

1.631g

1.019g

To determine the amount of sodium chloride (NaCl) required to produce 0.25 grams of silver chloride (AgCl), we need to use stoichiometry. The balanced chemical equation for the reaction between silver nitrate (AgNO3) and sodium chloride is as follows:

AgNO3 + NaCl -> AgCl + NaNO3

From the equation, we can see that the molar ratio between AgCl and NaCl is 1:1. This means that for every 1 mole of AgCl, we need 1 mole of NaCl.

To solve the problem, we follow these steps:

Step 1: Calculate the molar mass of AgCl
- The molar mass of AgCl is the sum of the molar masses of silver (Ag) and chlorine (Cl).
- Ag: Atomic weight = 107.87 g/mol
- Cl: Atomic weight = 35.45 g/mol
- AgCl: Molar mass = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol

Step 2: Convert the mass of AgCl to moles
- Given: Mass of AgCl = 0.25 g
- Moles of AgCl = Mass of AgCl / Molar mass of AgCl
= 0.25 g / 143.32 g/mol
≈ 0.0017446 mol

Step 3: Determine the amount of NaCl needed
- Since the molar ratio between AgCl and NaCl is 1:1, the moles of NaCl required will be the same as the moles of AgCl.
- Moles of NaCl needed = Moles of AgCl = 0.0017446 mol

Step 4: Convert the moles of NaCl to grams
- Given: Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol
- Mass of NaCl = Moles of NaCl x Molar mass of NaCl
= 0.0017446 mol x 58.44 g/mol
≈ 0.102 g

Therefore, the answer is option a) 0.102 grams of sodium chloride.