10.0 mL10.0 mL of 0.10 M0.10 M sodium chloride solution is mixed with 10.0 mL10.0 mL of 0.25 M0.25 M silver nitrate solution. After filtering and drying the product, 0.04 g0.04 g of silver chloride was produced.What is the percent yield from this reaction?
Question
10.0 mL10.0 mL of 0.10 M0.10 M sodium chloride solution is mixed with 10.0 mL10.0 mL of 0.25 M0.25 M silver nitrate solution. After filtering and drying the product, 0.04 g0.04 g of silver chloride was produced.What is the percent yield from this reaction?
Solution
To calculate the percent yield, we first need to determine the theoretical yield of the reaction.
The balanced chemical equation for the reaction is:
AgNO3 + NaCl -> AgCl + NaNO3
From the stoichiometry of the reaction, we can see that one mole of AgNO3 reacts with one mole of NaCl to produce one mole of AgCl.
First, we need to calculate the number of moles of each reactant:
Moles of NaCl = Volume (L) x Concentration (M) = 0.01 L x 0.10 M = 0.001 moles Moles of AgNO3 = Volume (L) x Concentration (M) = 0.01 L x 0.25 M = 0.0025 moles
The limiting reactant is NaCl because it has fewer moles. Therefore, the theoretical yield of AgCl is 0.001 moles.
Next, we convert moles of AgCl to grams using its molar mass (143.32 g/mol):
Theoretical yield = moles x molar mass = 0.001 moles x 143.32 g/mol = 0.14332 g
Finally, we calculate the percent yield:
Percent yield = (actual yield / theoretical yield) x 100% Percent yield = (0.04 g / 0.14332 g) x 100% = 27.9%
So, the percent yield of the reaction is 27.9%.
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