24.0 mL of a 2.4 M silver nitrate is mixed with 32.0 mL of 2.0 M sodium.
Calculate the mass of precipitae formed.
Calculate the concentration of all ions remaining in solution after the precipitate forms.
I don't know how to start this or where to go with it...
write a balanced equation.
figure the moles of each chem you have.
Determine the limiting reageant.
from the limiting reageant, determine the mass formed.
and the leftover excess reactant yields all the ions....
Sodium...? Sodium what?
Anyway, I'll just provide steps:
(1) Write the balanced chemical reaction. Actually I'm not sure with the reaction of this, because of the 'sodium'. Normally it would be Sodium Hydroxide, Sodium Chloride solution, etc.
(2) Calculate the moles of each from the molarity formula:
M = n/V
where M is the molarity (mol/L), n is the number of moles and V is the volume in Liters. You have to solve for n.
(3) Determine the limiting reactant. Make mol product / mol reaction ratios based from the chemical reaction, to solve for the moles of precipitate formed. Whichever of the two produces smaller number of moles of precipitate, that reactant is limiting, and you'll base from it the moles of precipitate formed.
(4) Multiply the number of moles you got in #3 by the molar mass of the precipitate to get its mass.
hope this helps~ `u`
To calculate the mass of precipitate formed, we first need to identify the chemical reaction that occurs between silver nitrate (AgNO3) and sodium chloride (NaCl). In this case, the reaction forms a precipitate called silver chloride (AgCl) and generates sodium nitrate (NaNO3) as a byproduct. The balanced chemical equation for this reaction is:
AgNO3 + NaCl -> AgCl + NaNO3
To find the mass of the precipitate formed, we need to determine the limiting reactant. The limiting reactant is the one that is completely consumed and thus determines the amount of product that can be formed.
Step 1: Calculate the number of moles of silver nitrate and sodium chloride:
Number of moles of AgNO3 = volume (in L) * concentration (in mol/L)
Number of moles of AgNO3 = 24.0 mL * (1 L / 1000 mL) * 2.4 mol/L
Number of moles of AgNO3 = 0.0576 mol
Number of moles of NaCl = volume (in L) * concentration (in mol/L)
Number of moles of NaCl = 32.0 mL * (1 L / 1000 mL) * 2.0 mol/L
Number of moles of NaCl = 0.064 mol
Step 2: Determine the stoichiometric ratio of AgNO3 to AgCl from the balanced equation.
From the balanced equation, we can see that 1 mole of AgNO3 reacts with 1 mole of AgCl.
Therefore, the number of moles of AgCl that can be formed = 0.0576 mol
Step 3: Calculate the mass of AgCl formed using its molar mass.
Molar mass of AgCl = atomic mass of Ag + atomic mass of Cl
Molar mass of AgCl = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol
Mass of AgCl formed = number of moles of AgCl * molar mass of AgCl
Mass of AgCl formed = 0.0576 mol * 143.32 g/mol
Mass of AgCl formed ≈ 8.24 g (rounded to two decimal places)
Therefore, the mass of silver chloride precipitate formed is approximately 8.24 grams.
To calculate the concentration of remaining ions in the solution after the precipitate forms, we need to consider the dissociation of the remaining compounds.
Step 4: Determine the ions present in solution after the reaction.
After the reaction, silver chloride (AgCl) precipitates, while sodium nitrate (NaNO3) remains in solution. Both compounds dissociate in aqueous solution.
Thus, the ions remaining in solution are sodium (Na+) and nitrate (NO3-) ions.
However, the reaction also involves the formation of sodium chloride (NaCl) as a byproduct, which is not included in the final solution.
Step 5: Calculate the concentration of sodium (Na+) ions.
Since the initial reaction mixture contained both sodium chloride and sodium nitrate, we need to calculate the total amount of sodium ions present before the reaction.
Number of moles of Na+ ions before the reaction = Number of moles of NaCl + Number of moles of NaNO3
Number of moles of Na+ ions before the reaction = 0.064 mol + (32.0 mL * (1 L / 1000 mL) * 2.0 mol/L)
Number of moles of Na+ ions before the reaction ≈ 0.064 mol + 0.064 mol ≈ 0.128 mol
Once we determine the moles of sodium ions present initially, we can calculate the concentration of sodium ions remaining after the reaction.
Concentration of sodium (Na+) ions = moles of Na+ ions remaining / total volume of solution
Concentration of sodium (Na+) ions = (moles of Na+ ions before the reaction - moles of Na+ ions used in the reaction) / total volume of solution
Concentration of sodium (Na+) ions = (0.128 mol - 0.064 mol) / (24.0 mL + 32.0 mL) * (1 L / 1000 mL)
Concentration of sodium (Na+) ions ≈ 0.002 mol/L
Step 6: Calculate the concentration of nitrate (NO3-) ions.
Since sodium nitrate (NaNO3) completely dissociates in solution, the concentration of nitrate ions remaining is the same as the initial concentration.
Concentration of nitrate (NO3-) ions ≈ 2.0 mol/L
Therefore, the concentration of remaining ions in solution after the precipitate forms is approximately 0.002 mol/L for sodium (Na+) ions and 2.0 mol/L for nitrate (NO3-) ions.