find the concentration of nitrate ions in the final reaction mixture. The percent yield of the experiment is 79.07%.The mass of the dried precipitate is 0.17g.The theoretical mass of precipitate produced in this reaction 0.215g.Also find the number of moles of precipitate produced in this reaction. CaCl2+2AgNo3 arrow Ca(No3)2(aq)+ClAg(s)
First, we need to calculate the actual mass of precipitate produced in the experiment:
Actual mass = 0.17g
Next, we can calculate the concentration of nitrate ions in the final reaction mixture using the actual mass of precipitate produced and the percent yield:
Percent yield = (Actual mass / Theoretical mass) x 100
79.07% = (0.17g / Theoretical mass) x 100
Theoretical mass = 0.17g / (79.07/100)
Theoretical mass = 0.215 g
So, the concentration of nitrate ions in the final reaction mixture can be calculated as:
Concentration = (Actual mass / Molar mass of precipitate) / Volume of final reaction mixture
Concentration = (0.17g / (2 x 169.87 g/mol)) / Volume
The molar mass of Ca(No3)2 = 2 x (14.01 + 3 x 16.00) = 2 x (14.01 + 48.00) = 2 x 62.01 = 124.02 g/mol
Concentration = (0.17g / (2 x 169.87 g/mol)) / Volume
Concentration = (0.17g / 339.74 g/mol) / Volume
Concentration = 0.0012508 mol/L / Volume
Lastly, we can calculate the number of moles of precipitate produced in the reaction:
Moles of precipitate = Mass of precipitate / Molar mass of precipitate
Moles of precipitate = 0.17g / 339.74 g/mol
We have now found the concentration of nitrate ions in the final reaction mixture and the number of moles of precipitate produced in this reaction.