A 0.050 M solution of silver nitrate (AgNO3) is added to a solution containing excess sodium chloride (NaCl). Calculate the mass of silver chloride (AgCl) precipitated when 50.0 mL of the AgNO3 solution is mixed with the NaCl solution
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To solve this problem, we need to determine the limiting reagent, which is the reactant that is completely consumed and determines the amount of product formed.
The balanced equation for the reaction is:
AgNO3 + NaCl -> AgCl + NaNO3
First, we need to convert the volume of the AgNO3 solution to moles using the concentration:
0.050 M = 0.050 moles/L
50.0 mL = 0.050 L
0.050 L * 0.050 moles/L = 0.0025 moles of AgNO3
Next, we need to determine the moles of NaCl:
Since NaCl is in excess, we can assume that all of the AgNO3 will react with NaCl to form AgCl. Therefore, the moles of AgNO3 will be the same as the moles of AgCl formed.
Now, we can calculate the mass of AgCl using its molar mass:
AgCl molar mass = 107.87 g/mol
Mass of AgCl = moles of AgCl * molar mass of AgCl
Mass of AgCl = 0.0025 moles * 107.87 g/mol
Mass of AgCl = 0.2697 g
Therefore, the mass of AgCl precipitated when 50.0 mL of the AgNO3 solution is mixed with the NaCl solution is 0.2697 grams.
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