I used 15.0mL of a 0.050 mol/L of calcium chloride, and 20,0mL of a 0.10 mol/L of silver nitrate. The balanced equation is CaCl2+2AgNO3 arrow Ca(No3)2+2AgCl.The mass of the filter paper is 1.58g. 1)Find the number of moles of precipitate produced in this reaction ? 2)Find the theoretical mass of precipiatate produced in this reaction.3)The mass of the precipitate of the dried precipitate was 0.17g. 4)Calculate the percent yield of this experiment.5)Find The concentration of nitrate ions in the final reaction mixture.
1) To find the number of moles of precipitate produced, we need to determine the limiting reactant.
The molar ratio between calcium chloride (CaCl2) and silver chloride (AgCl) is 1:2.
Moles of calcium chloride used = volume (in L) x concentration (mol/L)
Moles of calcium chloride used = 0.015 L x 0.050 mol/L = 0.00075 mol
Moles of silver nitrate used = volume (in L) x concentration (mol/L)
Moles of silver nitrate used = 0.020 L x 0.10 mol/L = 0.002 mol
According to the balanced equation, the mole ratio between calcium chloride and silver chloride is 1:2. So, for every mole of calcium chloride used, we would expect 2 moles of silver chloride to be produced.
Since we have 0.00075 mol of calcium chloride, this would produce 2 x 0.00075 = 0.0015 mol of silver chloride.
2) To find the theoretical mass of the precipitate produced, we need to know the molar mass of silver chloride (AgCl).
The molar mass of AgCl = atomic mass of Ag + atomic mass of Cl = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol
The theoretical mass of the precipitate produced = moles of silver chloride produced x molar mass of AgCl
The theoretical mass of the precipitate produced = 0.0015 mol x 143.32 g/mol = 0.21498 g ≈ 0.215 g
3) The mass of the dried precipitate was given as 0.17 g.
4) To calculate the percent yield, we need to divide the actual mass of the precipitate obtained (0.17 g) by the theoretical mass of the precipitate (0.215 g) and multiply by 100.
Percent yield = (actual mass of precipitate / theoretical mass of precipitate) x 100
Percent yield = (0.17 g / 0.215 g) x 100 = 79.07%
5) The concentration of nitrate ions in the final reaction mixture can be determined by stoichiometry. In the balanced equation, the mole ratio between silver nitrate (AgNO3) and nitrate ions (NO3-) is 2:2.
The moles of nitrate ions = 2 x moles of silver nitrate used = 2 x 0.002 mol = 0.004 mol
The volume of the final reaction mixture is the sum of the volumes of calcium chloride and silver nitrate used.
Volume of final reaction mixture = 0.015 L + 0.020 L = 0.035 L
The concentration of nitrate ions = moles of nitrate ions / volume of final reaction mixture
The concentration of nitrate ions = 0.004 mol / 0.035 L = 0.114 mol/L