A 0.8612g sample of a mixture consisting of NaBr,NaI and NaNO3 was analysed by adding AgNO3 to precipitate the Br^- and I^-,yielding 1.0186g mixture of AgBr and AgI.The precipitate was then heated in a stream of Cl2, converting it to 0.7125g of AgCl. Calculate the %w/w NaNO3 in the sample
30.72%
29.42%
30.72%
Eq2 and 3 can not be solved by simultaneously eq
To calculate the percentage by weight (w/w) of NaNO3 in the sample, we need to determine the mass of NaNO3 present in the mixture. Here's how we can solve the problem step by step:
1. Determine the mass of AgBr and AgI that were formed when AgNO3 was added to the mixture.
- Given: Mass of AgBr + AgI = 1.0186g
2. Calculate the mass of Cl that was released when the precipitate (AgBr and AgI) was heated with Cl2.
- Given: Mass of AgCl = 0.7125g
3. Determine the moles of AgBr and AgI that were formed when AgNO3 was added to the mixture. We can use the molecular weights of AgBr and AgI to convert the masses to moles.
- Molecular weight of AgBr = 187.77 g/mol
- Molecular weight of AgI = 234.77 g/mol
4. Calculate the moles of Cl that were released when the precipitate was heated with Cl2. We can use the molecular weight of AgCl to convert the mass of AgCl to moles.
- Molecular weight of AgCl = 143.32 g/mol
5. Use stoichiometry to determine the moles of NaNO3 that were originally present. Since NaBr and NaI react with AgNO3 in a 1:1 ratio, we can assume that the moles of NaBr and NaI are equal to the moles of AgBr and AgI that were formed.
6. Calculate the mass of NaNO3 using its molecular weight and the number of moles obtained in step 5.
7. Finally, calculate the percentage by weight of NaNO3 in the sample using the mass of NaNO3 and the total mass of the mixture.
Here are the calculations in more detail:
1. Mass of AgBr + AgI = 1.0186g
2. Mass of Cl released = Mass of AgCl = 0.7125g
3. Moles of AgBr = mass of AgBr / molecular weight of AgBr
Moles of AgI = mass of AgI / molecular weight of AgI
4. Moles of Cl = mass of AgCl / molecular weight of AgCl
5. Moles of NaBr and NaI = Moles of AgBr and AgI (assuming 1:1 stoichiometry)
6. Moles of NaNO3 = Moles of NaBr + Moles of NaI
7. Mass of NaNO3 = Moles of NaNO3 x molecular weight of NaNO3
8. Percentage by weight of NaNO3 = (Mass of NaNO3 / Total mass of sample) x 100
By following these steps and performing the calculations, you should be able to determine the percentage by weight of NaNO3 in the sample.
Three equations and three unknowns but the way they are set up makes it as if it were two equations and two unknowns.
Let X = mass NaX
and Y = mass NaI
and Z = mass NaNO3
(mm stands for molar mass)
---------------------------
eqn 1 is Z + Y + Z = 0.8612g
eqn 2 is mass AgBr from NaBr + mass AgI from NaI = mass AgBr + mass AgCl. This must be set up in terms of X and Y which is
eqn 2
X(mmAgBr/mmNaBr) + Y(mmAgI/mmNaI) = 1.0186g
eqn 3 is mass AgBr converted to AgCl + mass AgI converted to AgCl = mass AgCl and that set up in terms of X and Y is
eqn 3
X(mmAgCl/mmNaBr) | Y(mmAgCl/mmNaI) = 0.7125g
Solve 2 and 3 simultaneously to obtain X and Y, then plug X and Y into equation 1 to obtain Z (in grams).
Then %Z = (mass Z/mass sample)*100 = ?
Post your work if you have trouble.