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.
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)
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