What mass of silver chloride can be produced from 1.25 L of a 0.264 M solution of silver nitrate?
To find the mass of silver chloride that can be produced from the given solution of silver nitrate, you need to follow these steps:
Step 1: Write the balanced chemical equation for the reaction that occurs between silver nitrate (AgNO3) and sodium chloride (NaCl) to produce silver chloride (AgCl).
AgNO3 + NaCl -> AgCl + NaNO3
Step 2: Determine the molar ratio between silver nitrate (AgNO3) and silver chloride (AgCl) from the balanced equation. In this case, the molar ratio is 1:1, meaning that one mole of silver nitrate produces one mole of silver chloride.
Step 3: Calculate the number of moles of silver nitrate (AgNO3) from the given volume and concentration using the formula:
moles = volume (in liters) x concentration (in Molarity)
moles = 1.25 L x 0.264 M = 0.33 moles AgNO3
Step 4: Since the molar ratio between AgNO3 and AgCl is 1:1, the number of moles of AgCl produced is also 0.33 moles.
Step 5: Calculate the mass of AgCl produced using the molar mass of AgCl, which is 143.32 g/mol:
mass = moles x molar mass
mass = 0.33 moles x 143.32 g/mol = 47.29 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 47.29 grams.