0 votes
67 views
in Chemical reactions by (550 points)
A chemistry student needs to determine the precipitate formed when 50 mL of 0.2 M silver nitrate solution is mixed with 75 mL of 0.1 M potassium chloride solution. What is the mass of the precipitate formed and what is the percent yield of the reaction?

1 Answer

0 votes
by (290 points)
To determine the mass of the precipitate formed, we first need to identify the reaction taking place. The reaction between silver nitrate  AgNO3  and potassium chloride  KCl  is a double displacement reaction, which forms silver chloride  AgCl  as a precipitate and potassium nitrate  KNO3  in solution:AgNO3 aq  + KCl aq   AgCl s  + KNO3 aq Next, we need to determine the limiting reactant. We can do this by calculating the moles of each reactant and comparing their mole ratios.Moles of AgNO3 =  volume  concentration  =  50 mL  0.2 M  = 10 mmolMoles of KCl =  volume  concentration  =  75 mL  0.1 M  = 7.5 mmolThe stoichiometry of the reaction is 1:1, so we can see that KCl is the limiting reactant.Now, we can calculate the moles of AgCl formed, which will be equal to the moles of the limiting reactant  KCl :Moles of AgCl = 7.5 mmolTo find the mass of the precipitate, we need to convert moles of AgCl to grams using the molar mass of AgCl  143.32 g/mol :Mass of AgCl =  moles  molar mass  =  7.5 mmol  143.32 g/mol  = 1074.9 mgSo, the mass of the precipitate formed is 1074.9 mg.To calculate the percent yield, we need to know the theoretical yield and the actual yield. In this case, we don't have information about the actual yield, so we cannot calculate the percent yield. If the actual yield was provided, the percent yield could be calculated using the formula:Percent yield =  actual yield / theoretical yield   100%

Related questions

Welcome to Sarvan Science Q&A, where you can ask questions and receive answers from other members of the community.
...