A 10.00 gram solid sample consisting of a mixture of sodium chloride and potassium sulfate is dissolved in 100.0 mL of water is combined with another 200.0 mL solution that contains excess aqueous lead(II) nitrate. Upon mixing the two solutions, a solid precipitate forms (it is a mixture of two ionic solids.) It is filtered, dried, and found to have a dry mass of 21.75 grams. Determine the mass percent of sodium chloride in the original solid sample mixture of sodium chloride and potassium sulfate.

Hi, Bob. But how to get (MMPbCl2/2*MMNaCl) and (MMPbSO4/MMK2SO4). Because it is confusing to me. Can you show the step more clearly. Thank you?

I just can do step 1, which is X=10-Y.
But step 2, it is bit confused. I can't get the meaning.

Two equations in two unknowns; solve simultaneously.

Let X = mass NaCl
and Y = mass K2SO4
-------------------
eqn 1 is X + Y = 10.0
eqn 2 is mass PbCl2 (from NaCl) + mass PbSO4(from K2SO4) = 21.75g.
eqn 2 in terms of X and Y is this

(X*MMPbCl2/2*MMNaCl) + (Y*MMPbSO4/MMK2SO4) = 21.75

Solve those two equations for X and Y, then
%NaCl = (Xgrams/10)*100 = ?

Well, it seems like we've got a classic chemistry problem here! Let's calculate the mass percent of sodium chloride in the original solid sample mixture.

First, let's find the mass of the precipitate by subtracting the initial mass of the filter paper from the dry mass of the precipitate. Unfortunately, I can't make a joke about this step – it's just straightforward math!

Mass of precipitate = 21.75 grams - (mass of filter paper)

Next, let's find the moles of lead(II) nitrate in the combined solutions using the concentration and volume of the lead(II) nitrate solution. Wait, let me think... how do I make a joke about moles? Alright, here it is:

Why was Avogadro so rich? Because he was a "mole"ti-millionaire! Get it? "Mole"-ti-millionaire? Ah, chemistry humor!

Anyhow, once we have the moles of lead(II) nitrate, we can use stoichiometry to find the moles of sodium chloride that reacted. Remember, sodium chloride and lead(II) nitrate are in a 1:2 ratio because of the balanced chemical equation. And no, the balanced equation isn't a tightrope-walking act between the two!

After calculating the moles of sodium chloride that reacted, convert it to grams. Then, divide the mass of sodium chloride by the mass of the original solid sample and multiply by 100 to get the mass percent.

Okay, this part is serious – no more jokes. Here's the equation to help you out:

Mass percent of sodium chloride = (mass of sodium chloride / mass of original solid sample) x 100.

I hope that helps! And remember, chemistry might be challenging, but humor makes it more chlorine... I mean, appealing!

To determine the mass percent of sodium chloride in the original solid sample mixture, we need to follow these steps:

1. Calculate the mass of the precipitate formed:
- The mass before the reaction is 10.00 grams.
- The mass after drying is 21.75 grams.
- Therefore, the mass of the precipitate formed is 21.75 grams - 10.00 grams = 11.75 grams.

2. Determine the mass of sodium chloride in the precipitate:
- Sodium chloride is one of the ionic solids in the mixture.
- The molar mass of sodium chloride is 58.44 g/mol.
- To find the mass of sodium chloride in the precipitate, we need to calculate the number of moles of sodium chloride by dividing the mass by the molar mass:
Moles of sodium chloride = mass of sodium chloride / molar mass of sodium chloride
= (mass percent of sodium chloride / 100) * mass of precipitate / molar mass of sodium chloride.

3. Calculate the mass of sodium chloride in the original solid sample:
- Since the precipitate is a mixture of sodium chloride and potassium sulfate, the total mass of the mixture is still 10.00 grams.
- The mass of sodium chloride in the original solid sample would be the same as the mass of sodium chloride in the precipitate.

4. Determine the mass percent of sodium chloride in the original solid sample:
- Mass percent of sodium chloride = (mass of sodium chloride in the original sample / mass of the original sample) * 100.

By following these steps and substituting the values, you can calculate the mass percent of sodium chloride in the original solid sample mixture of sodium chloride and potassium sulfate.

Thank you, I know how to do it.