Answer:
We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution
Explanation:
<u>Step 1:</u> Data given
Sample 1: The 1.15 M sample has a volume of 33.O mL
Sample 2: The 0.660 M sample has a volume of 59.0 mL
Molar mass of KBr = 119 g/mol
Molar mass of AgNO3 = 169.87 g/mol
<u>Step 2:</u> Calculate number of moles for both samples
Number of moles = Molarity * Volume
Sample 1: 1.15 M * 33 *10^-3 L = 0.03795 moles
Sample 2: 0.660 M *59*10^-3 L = 0.03894 moles
Total mol KBr = 0.03795 + 0.03894 = 0.07689 moles
<u>Step 3:</u> Calculate total mass
mass = Number of moles * Molar mass
mass = 0.07689 moles * 119 g/moles = 9.15 grams ( in 55mL)
<u>Step 4</u>: Calculate moles of AgBr
AgNO3 reacts with KBr
KBr(aq) + AgNO3(aq) → AgBr(s) + KNO3(aq)
1 mole of KBr consumed, needs 1 mole of AgNO3 to produce 1 mole of AgBr and 1 mole of KNO3
So 0.07689 moles of KBr wll need 0.07689 moles of AgNO3
<u>Step 5:</u> Calculate mass of silver nitrate
mass of AgNO3 = Moles of AgNO3 * Molar mass of AgNO3
mass of AgNO3 = 0.07689 moles * 169.87 g/mol = 13.06 grams
We need 13.06 grams of silver nitrate to precipitate out silver bromide in the final solution
