A 1.91-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 0.44 g. What is the percentage by mass of NaNO3 in the original mixture?
I know I have to write the equation and balance it then find the moles of Ag2S (since NO3 does not precipitate), but I'm not sure where to go from there.
Cheers
Right so far. Find mols Ag2S, convert to mols NaNO3, convert to g NaNO3 and stick in the % formula.
%NaNO3 = (mass NaNO3/mass sample)*100 = ?
How do I convert mols Ag2S to mols NaNO3?
wes, use the balanced equation and the coefficients of Ag2S and NaNO2.
To determine the percentage by mass of NaNO3 in the original mixture, you can follow these steps:
1. Write and balance the equation:
AgNO3(aq) + Na2S(aq) -> Ag2S(s) + 2NaNO3(aq)
2. Calculate the molar mass of Ag2S:
Ag: 2 x 107.87 g/mol = 215.74 g/mol
S: 1 x 32.07 g/mol = 32.07 g/mol
Total molar mass of Ag2S: 215.74 + 32.07 = 247.81 g/mol
3. Determine the number of moles of Ag2S formed:
molar mass = mass / moles
moles = mass / molar mass
moles of Ag2S = 0.44 g / 247.81 g/mol = 0.0017752 mol
4. Use the balanced equation to find the moles of NaNO3:
From the balanced equation, we know that 1 mole of Ag2S precipitates for every 2 moles of NaNO3.
Therefore, moles of NaNO3 = 2 x moles of Ag2S = 2 x 0.0017752 mol = 0.0035504 mol
5. Calculate the mass of NaNO3:
mass = moles x molar mass
mass of NaNO3 = 0.0035504 mol x (22.99 g/mol + 14.01 g/mol + 3 x 16.00 g/mol) = 0.19 g
6. Determine the percentage by mass of NaNO3 in the original mixture:
percentage by mass = (mass of NaNO3 / mass of original mixture) x 100%
percentage by mass = (0.19 g / 1.91 g) x 100% = 9.95%
Therefore, the percentage by mass of NaNO3 in the original mixture is approximately 9.95%.