A 1.27-g sample of a mixture of AgNO3 and NaNO3 is treated with excess Na2S(aq). The precipitate is filtered off, dried and weighed. The dried precipitate weighs 0.46 g. What is the percentage by mass of NaNO3 in the original mixture?

The Na2S reacts with AgNO3 but not with NaNO3.

Na2S + 2AgNO3 ==> Ag2S + 2NaNO3
mols Ag2S = 0.46/molar mass Ag2S = ?
Using the coefficients in the balanced equation, convert mols Ag2S to mols AgNO3.
Since the mixture has a mass of 1.27 g, 1.27- mass AgNO3 from above = mass NaNO3.
Then % NaNO3 = (mass NaNO3/mass sample)*100 = ?

To find the percentage by mass of NaNO3 in the original mixture, we need to calculate the mass of NaNO3 in the precipitate and then compare it to the mass of the original mixture.

Here's how you can do it step by step:

1. Calculate the mass of AgNO3 in the precipitate:
- Start by assuming that all the AgNO3 in the original mixture is converted into precipitate.
- The molar mass of AgNO3 is 169.87 g/mol.
- Since the precipitate weighs 0.46 g, you can calculate the amount of AgNO3 using the following equation:
Amount of AgNO3 = (mass of precipitate / molar mass of AgNO3)

2. Calculate the mass of NaNO3 in the precipitate:
- Subtract the mass of AgNO3 in the precipitate from the total mass of the precipitate:
Mass of NaNO3 = Total mass of precipitate - Mass of AgNO3

3. Calculate the mass of NaNO3 in the original mixture:
- Divide the mass of NaNO3 in the precipitate by the mass of the original mixture and multiply by 100 to get the percentage.
Percentage of NaNO3 = (Mass of NaNO3 / Mass of original mixture) x 100

Now let's apply these steps to solve the problem:

Given:
Mass of the original mixture = 1.27 g
Mass of the precipitate = 0.46 g

Step 1: Calculate the mass of AgNO3 in the precipitate.
Molar mass of AgNO3 = 169.87 g/mol
Amount of AgNO3 = (0.46 g / 169.87 g/mol) = 0.00271 mol

Step 2: Calculate the mass of NaNO3 in the precipitate.
Mass of NaNO3 = 0.46 g - (0.00271 mol x 169.87 g/mol) = 0.46 g - 0.4599 g = 0.0001 g

Step 3: Calculate the percentage of NaNO3 in the original mixture.
Percentage of NaNO3 = (0.0001 g / 1.27 g) x 100 = 0.00787 x 100 = 0.787%

Therefore, the percentage by mass of NaNO3 in the original mixture is approximately 0.787%.