For the following reaction,determine the amount of agcl formed (collected in laboratory) when 10.24 g of kcl react with excess agno3 to produce agcl and kno3 with a 77.24% yield.
To determine the amount of AgCl formed, we need to use stoichiometry and the given information.
Step 1: Write the balanced chemical equation for the reaction:
2 AgNO3 + 2 KCl → 2 AgCl + 2 KNO3
According to the balanced equation, 2 moles of AgCl are produced for every 2 moles of KCl reacting.
Step 2: Calculate the number of moles of KCl:
Given mass of KCl = 10.24 g
Molar mass of KCl = 74.55 g/mol
Number of moles of KCl = (given mass of KCl) / (molar mass of KCl)
Number of moles of KCl = 10.24 g / 74.55 g/mol
Step 3: Determine the theoretical yield of AgCl:
Theoretical yield of AgCl = (number of moles of KCl) * (molar ratio of AgCl to KCl)
In the balanced equation, the molar ratio of AgCl to KCl is 2:2. Therefore, one mole of KCl produces one mole of AgCl.
Theoretical yield of AgCl = (number of moles of KCl) * 1
Theoretical yield of AgCl = number of moles of KCl
Step 4: Calculate the actual yield of AgCl:
Actual yield of AgCl = (theoretical yield of AgCl) * (percent yield)
Given percent yield = 77.24%
Actual yield of AgCl = (theoretical yield of AgCl) * (0.7724)
Step 5: Calculate the mass of AgCl formed:
Molar mass of AgCl = 143.32 g/mol
Mass of AgCl formed = (actual yield of AgCl) * (molar mass of AgCl)
Now, substitute the values into the equation and calculate:
Number of moles of KCl = 10.24 g / 74.55 g/mol
Number of moles of KCl = 0.1374 mol
Theoretical yield of AgCl = 0.1374 mol
Actual yield of AgCl = 0.1374 mol * 0.7724
Actual yield of AgCl = 0.1061 mol
Mass of AgCl formed = 0.1061 mol * 143.32 g/mol
Mass of AgCl formed = 15.20 g
Therefore, the amount of AgCl formed (collected in the laboratory) is 15.20 grams.