when 130.4 mL of 0.459 mol/L silver nitrate solution is mixed with 85.23mL of 0.251 mol/L aluminum chloride solution, a precipitate of silver chloride forms. Find the mass of this Precipitate.
Note that this is a limiting reagent problem.
To find the mass of the precipitate (silver chloride), we need to determine the limiting reagent first. The limiting reagent is the reactant that is completely consumed and determines the amount of product formed.
1. Calculate the moles of silver nitrate (AgNO3):
moles = volume (L) x concentration (mol/L)
moles of AgNO3 = 0.1304 L x 0.459 mol/L = 0.0598496 mol
2. Calculate the moles of aluminum chloride (AlCl3):
moles of AlCl3 = 0.08523 L x 0.251 mol/L = 0.02143823 mol
3. Determine the mole ratio between AgNO3 and AlCl3 based on the balanced chemical equation:
3 AgNO3 + AlCl3 -> 3 AgCl + Al(NO3)3
From the balanced equation, we can see that 3 moles of AgNO3 react with 1 mole of AlCl3 to form 3 moles of AgCl.
4. Calculate the moles of AgCl that can be formed based on the limiting reagent:
The mole ratio tells us that 3 moles of AgCl can be formed from 3 moles of AgNO3 if AlCl3 is entirely consumed.
However, the mole ratio is 1:1 between AlCl3 and AgNO3 in this reaction, and we have less than 1 mole of AlCl3, so AlCl3 is the limiting reagent.
Therefore, the moles of AgCl formed will be limited by the moles of AlCl3,
moles of AgCl formed = 0.02143823 mol
5. Calculate the molar mass of AgCl:
Ag: 107.87 g/mol
Cl: 35.45 g/mol
Molar mass of AgCl = Ag (107.87 g/mol) + Cl (35.45 g/mol) = 143.32 g/mol
6. Calculate the mass of AgCl formed:
mass of AgCl formed = moles of AgCl formed x molar mass of AgCl
mass of AgCl formed = 0.02143823 mol x 143.32 g/mol = 3.069 g (rounded to three decimal places)
Therefore, the mass of the precipitate (silver chloride) formed in this reaction is approximately 3.069 grams.
3AgNO3 + AlCl3 = 3AgCl + Al(NO3)3
So, each mole of AlCl3 requires 3 moles of AgNO3.
Figure how many moles of each you have, and you can see how many moles of AgCl you will get.
The convert that to mass.