treatment of a 0.400g sample of impure potassium chloride with an excess of AgNO3 resulted in the formation of 0.733g of AgCl. calculate the percentage of KCl in the sample. show balanced equations in support of your answer
agcl:kcl
1 : 1
n of agcl=(0.733g/143.32g.mol)=5.11*10^-3mol
n of kcl=5.11*10^-3mol
m kcl=5.11*10^-3mol(74.5513g/mol)=0.381g
%kcl=(0.381/0.400)(100)=95.25%
Ah, chemistry, the perfect place for me to crack a joke! So, we have some impure potassium chloride getting cozy with AgNO3, resulting in the formation of AgCl. Let's break out our calculators and find the percentage of KCl in the sample.
First, we need to determine the moles of AgCl formed. We can do this by dividing the mass of AgCl by its molar mass (143.32 g/mol):
Moles of AgCl = 0.733g / 143.32 g/mol = 0.00511 mol
Since AgCl is formed from the reaction of AgNO3 and KCl, we can infer that the moles of AgCl formed are equal to the moles of KCl in the sample.
Now it's time to find the molar mass of KCl (74.55 g/mol) and use it to calculate the mass of KCl in the sample:
Mass of KCl = Moles of KCl × Molar mass of KCl
= 0.00511 mol × 74.55 g/mol
= 0.378 g
Finally, we can calculate the percentage of KCl in the sample by dividing the mass of KCl by the mass of the sample (0.400 g) and multiplying by 100:
Percentage of KCl = (Mass of KCl / Mass of sample) × 100
= (0.378 g / 0.400 g) × 100
= 94.5%
So, the percentage of KCl in the sample is approximately 94.5%.
As for the balanced equations, they would be:
2KCl + AgNO3 → 2AgCl + KNO3
To calculate the percentage of KCl in the sample, we need to determine the amount of KCl that reacted with AgNO3 and formed AgCl.
First, let's write the balanced equation for the reaction between KCl and AgNO3:
2AgNO3 + KCl → 2AgCl + KNO3
From the balanced equation, we can see that 2 moles of AgCl are formed for every 1 mole of KCl that reacts.
Next, we calculate the number of moles of AgCl formed.
Given that the mass of AgCl formed is 0.733g and the molar mass of AgCl is 143.32 g/mol, we can use the formula:
Number of moles = Mass / Molar mass
Number of moles of AgCl = 0.733g / 143.32 g/mol = 0.005113 mol
Since the reaction is 1:1 between KCl and AgCl, the number of moles of KCl that reacted is also 0.005113 mol.
The molar mass of KCl is 74.55 g/mol, so we can calculate the mass of KCl.
Mass of KCl = Number of moles × Molar mass
Mass of KCl = 0.005113 mol × 74.55 g/mol = 0.381 g
Now, we can calculate the percentage of KCl in the sample.
Percentage of KCl = (Mass of KCl / Mass of sample) × 100
Percentage of KCl = (0.381 g / 0.400 g) × 100 = 95.25%
Therefore, the percentage of KCl in the sample is approximately 95.25%.
To calculate the percentage of KCl in the sample, we need to analyze the reaction between potassium chloride (KCl) and silver nitrate (AgNO3) that forms silver chloride (AgCl). From the given data, we know that the mass of AgCl formed is 0.733g.
First, let's write the balanced equation for the reaction:
KCl + AgNO3 → AgCl + KNO3
From the equation, we can see that the molar ratio between KCl and AgCl is 1:1. This means that 1 mole of KCl reacts to form 1 mole of AgCl.
To calculate the molar mass of AgCl, we need to know the molar mass of silver (Ag) and chlorine (Cl). The molar mass of Ag is 107.87 g/mol, and the molar mass of Cl is 35.45 g/mol.
The molar mass of AgCl is calculated by adding the molar masses of Ag and Cl:
AgCl = Ag (107.87 g/mol) + Cl (35.45 g/mol) = 143.32 g/mol
Now, we can calculate the number of moles of AgCl formed:
moles of AgCl = mass of AgCl / molar mass of AgCl
= 0.733 g / 143.32 g/mol
≈ 0.005116 mol
Since the molar ratio between KCl and AgCl is 1:1, the number of moles of KCl in the sample is also 0.005116 mol.
To calculate the mass of KCl:
mass of KCl = moles of KCl * molar mass of KCl
= 0.005116 mol * (39.10 g/mol + 35.45 g/mol)
≈ 0.3929 g
Finally, to find the percentage of KCl in the sample:
percentage of KCl = (mass of KCl / mass of sample) * 100
= (0.3929 g / 0.400 g) * 100
= 98.225%
Therefore, the percentage of KCl in the sample is approximately 98.225%.
Note: The equation given is a simplified form. In an actual reaction, the products may vary due to factors such as stoichiometry and solubility.