10 grams of sodium chloride is treated with excess silver nitrate AgNO3 + NaCl =AgCl +NaNO3.. HOW MUCH IS PERCIPITATED

A worked example of a stoichiometry problem.

http://www.jiskha.com/science/chemistry/stoichiometry.html

To determine how much precipitate is formed in the reaction between 10 grams of sodium chloride (NaCl) and excess silver nitrate (AgNO3), we need to understand the stoichiometry of the reaction.

The balanced chemical equation for the reaction is:
AgNO3 + NaCl → AgCl + NaNO3

From the equation, you can see that one mole of AgNO3 reacts with one mole of NaCl to produce one mole of AgCl. We need to find the number of moles of NaCl reacting to determine the amount of AgCl precipitated.

To calculate the number of moles of NaCl, we need to know the molecular weight of NaCl. The atomic masses of sodium (Na) and chlorine (Cl) are approximately 23 g/mol and 35.5 g/mol, respectively.

The molecular weight of NaCl is:
Na (23 g/mol) + Cl (35.5 g/mol) = 58.5 g/mol

Using the molecular weight, we can calculate the number of moles of NaCl:
Number of moles = Mass / Molecular weight
Number of moles = 10 g / 58.5 g/mol ≈ 0.171 moles

Since the stoichiometry of the reaction is 1:1 between NaCl and AgCl, we can conclude that approximately 0.171 moles of AgCl will be precipitated.

If you want to calculate the mass of AgCl precipitated, you can use the molar mass of AgCl. The atomic masses of silver (Ag) and chlorine (Cl) are approximately 107.87 g/mol and 35.5 g/mol, respectively.

The molecular weight of AgCl is:
Ag (107.87 g/mol) + Cl (35.5 g/mol) = 143.37 g/mol

To calculate the mass of AgCl, we multiply the number of moles by the molar mass:
Mass of AgCl = Number of moles × Molecular weight
Mass of AgCl = 0.171 moles × 143.37 g/mol ≈ 24.52 grams

Therefore, approximately 24.52 grams of AgCl will be precipitated in the reaction.