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 = ?
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?
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 = ?
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.