A 6.70 g sample of a binary mixture of silver (I) nitrate and sodium nitrate is completely
dissolved in water to make a solution. Aqueous barium chloride is then added to this solution
in excess to form 3.60 g of silver chloride solid. What is the percent by mass of silver nitrate
in the initial binary mixture?
Ag^+ + Cl^- ==> AgCl
mols AgCl = grams/molar mass =?
? mol AgCl = mols Cl^- = mols Ag^+ = mols AgNO3.
g AgNO3 = mols AgNO3 x molar mass AgNO3
%AgNO3 = (g AgNO3/mass sample)*100 = ?
Thank you!
To find the percent by mass of silver nitrate in the initial binary mixture, we need to determine the amount of silver nitrate and sodium nitrate present in the mixture.
Let's assume that x grams of silver nitrate is present in the mixture. Therefore, the mass of sodium nitrate would be (6.70 - x) grams.
Now, let's calculate the amount of silver chloride formed when barium chloride is added in excess. The reaction between silver nitrate and barium chloride can be represented as follows:
2AgNO3(aq) + BaCl2(aq) → 2AgCl(s) + Ba(NO3)2(aq)
From the balanced equation, we see that 2 moles of AgNO3 react to form 2 moles of AgCl. Therefore, the molar mass of AgNO3 is equal to the molar mass of AgCl.
Using the molar masses of Ag and Cl (107.87 g/mol and 35.45 g/mol, respectively), we can calculate the amount of AgCl formed:
Amount of AgCl = (mass of AgCl) / (molar mass of AgCl)
Given that the mass of AgCl formed is 3.60 g, we can substitute these values into the equation:
Amount of AgCl = 3.60 g / (107.87 g/mol)
Now, let's calculate the amount of silver nitrate by converting the amount of AgCl to moles:
Amount of AgNO3 = (Amount of AgCl) / (moles of AgNO3 produced per mole of AgCl)
From the balanced equation, we know that 2 moles of AgNO3 are required to produce 2 moles of AgCl. Therefore:
Amount of AgNO3 = (Amount of AgCl) / 2
Now we can substitute the calculated value for the amount of AgCl:
Amount of AgNO3 = (3.60 g / 107.87 g/mol) / 2
Next, let's calculate the mass of AgNO3:
Mass of AgNO3 = (Amount of AgNO3) * (molar mass of AgNO3)
Mass of AgNO3 = [(3.60 g / 107.87 g/mol) / 2] * (169.87 g/mol)
Finally, let's calculate the percent by mass of silver nitrate in the initial binary mixture:
Percent by mass of AgNO3 = (Mass of AgNO3 / total mass of mixture) * 100
Percent by mass of AgNO3 = (Mass of AgNO3 / 6.70 g) * 100
Substituting the calculated value for the mass of AgNO3:
Percent by mass of AgNO3 = [(3.60 g / 107.87 g/mol) / 2] * (169.87 g/mol) / 6.70 g * 100
Calculating this expression will give you the answer.