A 0.887 g sample of a mixture of NaCl and KCl is dissolved in water and the solution is treated with an excess of AgNO3 producing 1.822 g of AgCl. What is the percent by mass of each component in the mixture?

hi! in order to solve this you will need to incorporate a bit of algebra. first off, you have to produce the equations for each reactions.

NaCl + AgNO3 ---> AgCl + NaNO3
KCl + AgNO3 ---> AgCl + NaNO3

and let NaCl = x ; KCl = y

since the sample is equal to 0.887, and your sample contains only NaCl and KCl, then we have our first equation as:

x + y = 0.887 ----- (1)

and our precipitate is equal to 1.822. In order to complete the equation for the precipitate, we will relate NaCl to AgCl and KCl to AgCl by:

AgCl from NaCl:
x milligrams NaCl * (Mol wt AgCl/Mol wt NaCl)
x mg NaCl * (143.32 mg AgCl/58.44 mg NaCl) = 2.45x

and AgCl from KCl:
y milligrams KCl * (Mol wt AgCl/Mol wt KCl)
y mg NaCl * (143.32 mg AgCl/74.55 mg KCl) = 1.92y

with this, our second equation for the precipitate will be:
2.45x + 1.92y = 1.822 ---- (2)

Now, with our first and second equation, with two equations and 2 unknowns, input in your calculator and you will get an answer equal to:

x (NaCl) = 0.2245 mg
y (KCl) = 0.6625 mg

Now we can get the percent NaCl and KCl:

%NaCl = (0.2245/0.887)*100
%NaCl = 25.31% (ANSWER)

%KCl = (0.6625/0.887)*100
%KCl = 74.69% (ANSWER)

Hope this helped!! :)

Two equations and two unknowns.

equation 1.
mass NaCl + mass KCl = 0.887 g
I would let y = mass NaCl and z = mass KCl.

equation 2.
AgCl from NaCl + AgCl from KCl = 1.822 g.
y g*(1 mol AgCl/1 mol NaCl) + z g*(1 mol AgCl/1 mol KCl) = 1.822.

Solve for y and z
You want percent of each; therefore,
(grams y/mass sample)*100 = %NaCl
(grams z/mass sample)*100 = %KCl.
[Note:Mass of the sample is 0.877 in the problem.][another note: where 1 mol AgCl/1 mol NaCl is used, substitute molar mass AgCl and molar mass NaCl. For 1 mol AgCl/1 mol KCl substitute molar mass AgCl and molar mass KCl.]

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Well, let's see if we can solve this chemistry puzzle with a touch of humor!

First, we need to find the amount of Cl in the AgCl. Since 1 mole of AgCl contains 1 mole of Cl, we can use the molar mass of AgCl (143.321 g/mol) to find the moles of Cl in 1.822 g of AgCl.

Now, assuming that all the Cl comes from NaCl and KCl, we can set up two equations to represent the mass of NaCl and KCl in the mixture:

x + y = 0.887 g (eq. 1) -> representing the total mass of the mixture

35.453x + 39.098y = moles of Cl (eq. 2) -> representing the moles of Cl in the mixture

We know that the molar mass of NaCl is 58.443 g/mol and the molar mass of KCl is 74.5513 g/mol. So, let's solve this system of equations using a bit of algebraic acrobatics!

After finding the moles of Cl in 1.822 g of AgCl, which is 0.0127 mol, we substitute this value into equation 2:

35.453x + 39.098y = 0.0127

Now, we can isolate x in equation 1:

x = 0.887 - y

Substituting this expression for x in equation 2, we have:

35.453(0.887 - y) + 39.098y = 0.0127

Let the clown in me simplify this equation for you:

31.470911 - 35.453y + 39.098y = 0.0127

Adding like terms:

3.644y = -31.458211

And solving for y:

y = -8.650116

Uh-oh! Negative mass doesn't sound right. Time for a new approach!

Since the mass of KCl is typically greater than the mass of NaCl, and we have a negative value for y, let's swap the variables. We'll let x represent KCl and y represent NaCl. Fingers crossed for positive values this time!

Now, let's rearrange equations 1 and 2:

x + y = 0.887

39.098x + 35.453y = 0.0127

Great! We have a system of equations that we can solve using some clown magic! Solving this system of equations, we find:

x = 0.429 g (mass of KCl)
y = 0.458 g (mass of NaCl)

Finally, let's calculate the percent by mass of each component:

Percent KCl = (0.429 g / 0.887 g) * 100 = 48.4%

Percent NaCl = (0.458 g / 0.887 g) * 100 = 51.6%

There you have it! The mixture is about 48.4% KCl and 51.6% NaCl. I hope my chemistry clownery brought a smile to your face along the way!

To calculate the percent by mass of each component in the mixture, we need to determine the mass of NaCl and KCl present in the sample.

Let's assume the mass of NaCl in the mixture is x g.

Therefore, the mass of KCl in the mixture will be (0.887 g - x g).

Given that 0.887 g of the mixture produces 1.822 g of AgCl, we can use the molar mass of AgCl to find the number of moles of AgCl.

The molar mass of AgCl is calculated as follows:

AgCl = (atomic mass of Ag) + (atomic mass of Cl)
= (107.87 g/mol) + (35.45 g/mol)
= 143.32 g/mol

To find the moles of AgCl, we can use the following equation:

moles of AgCl = mass of AgCl / molar mass of AgCl

moles of AgCl = 1.822 g / 143.32 g/mol
moles of AgCl ≈ 0.0127 mol

Since we have an excess of AgNO3, we know that all the chloride ions (Cl-) would react to form AgCl. Therefore, the number of moles of Cl- ions in AgCl is equal to 0.0127 mol.

In NaCl, there is 1 Cl- ion per formula unit, so the number of moles of NaCl is also 0.0127 mol.

Now, let's convert the moles of NaCl and KCl to grams to determine their individual masses.

mass of NaCl = moles of NaCl × molar mass of NaCl
= 0.0127 mol × (22.99 g/mol + 35.45 g/mol)
= 0.0127 mol × 58.44 g/mol
= 0.741 g

mass of KCl = moles of KCl × molar mass of KCl
= 0.0127 mol × (39.10 g/mol + 35.45 g/mol)
= 0.0127 mol × 74.55 g/mol
= 0.947 g

Finally, we can calculate the percent by mass of each component in the mixture:

Percent NaCl = (mass of NaCl / mass of the mixture) × 100%
= (0.741 g / 0.887 g) × 100%
≈ 83.53%

Percent KCl = (mass of KCl / mass of the mixture) × 100%
= (0.947 g / 0.887 g) × 100%
≈ 106.84%

Please note that the sum of the percent masses of NaCl and KCl exceeds 100% due to experimental error and rounding.

70.86