What mass of silver chloride can be produced from 1.25 L of a 0.264 M solution of silver nitrate?

A sample problem worked. Remember, moles = M x L.http://www.jiskha.com/science/chemistry/stoichiometry.html

To find the mass of silver chloride produced, we need to use the concept of stoichiometry and calculate the amount of silver chloride formed based on the reaction between silver nitrate and chloride ions.

First, let's write the balanced chemical equation for the reaction:

AgNO3 + NaCl -> AgCl + NaNO3

From the balanced equation, we can see that one mole of silver nitrate (AgNO3) reacts with one mole of chloride ions (Cl-) to produce one mole of silver chloride (AgCl).

To determine the amount of silver chloride produced, we can use the formula:

Amount (in moles) = concentration (in moles per liter) x volume (in liters)

Given:
Concentration of silver nitrate solution = 0.264 M
Volume of silver nitrate solution = 1.25 L

Using the formula, we can calculate the moles of silver nitrate:

Moles of AgNO3 = concentration x volume
= 0.264 M x 1.25 L
= 0.33 moles

Since the stoichiometry of the reaction is 1:1, the moles of silver chloride formed will be equal to the moles of silver nitrate.

Therefore, the mass of silver chloride can be calculated using the molar mass of AgCl (143.32 g/mol):

Mass of AgCl = moles of AgCl x molar mass of AgCl
= 0.33 moles x 143.32 g/mol
= 47.26 g

Therefore, the mass of silver chloride that can be produced from 1.25 L of a 0.264 M solution of silver nitrate is approximately 47.26 grams.