In a chemical analysis, 3.40g of silver nitrate in solution reacted with excess sodium chloride to produce 2.81g of precipitate. What is the percent yield? Please show work. Thanks
Here is a worked example.
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http://www.jiskha.com/science/chemistry/stoichiometry.html
To calculate the percent yield, we need to compare the actual yield (the amount of product obtained) with the theoretical yield (the amount of product that should have been obtained according to stoichiometry).
First, we need to determine the stoichiometry of the reaction between silver nitrate (AgNO3) and sodium chloride (NaCl). The balanced equation for the reaction is:
AgNO3 + NaCl -> AgCl + NaNO3
From the equation, we see that 1 mole of silver nitrate reacts with 1 mole of sodium chloride to produce 1 mole of silver chloride.
Next, we need to determine the moles of silver chloride produced from the given mass:
Given:
Mass of silver nitrate = 3.40g
Mass of precipitate (silver chloride) = 2.81g
To find the moles of silver chloride, we divide the mass by its molar mass. The molar mass of silver chloride (AgCl) is 143.32 g/mol.
Moles of silver chloride = (2.81g) / (143.32 g/mol) = 0.0196 mol
Since the reaction is 1:1 between silver nitrate and silver chloride, the moles of silver nitrate consumed in the reaction are also 0.0196 mol.
Now, we calculate the theoretical yield of silver chloride. The molar mass of silver nitrate (AgNO3) is 169.87 g/mol.
Theoretical yield of silver chloride = (0.0196 mol) x (143.32 g/mol) = 2.81g
Comparing the actual yield and the theoretical yield:
Actual yield = 2.81g
Theoretical yield = 2.81g
Finally, we can calculate the percent yield using the formula:
Percent yield = (Actual yield / Theoretical yield) x 100
Percent yield = (2.81g / 2.81g) x 100 = 100%
Therefore, the percent yield is 100%.