calculate the mass of solid NaCl that must be added to 1.5L of 0.100M AgNO3 solution to precipitate all the Ag+ ions in the form of AgCl?
sdvdb
8.78g
To calculate the mass of solid NaCl needed to precipitate all the Ag+ ions in the form of AgCl, we need to use stoichiometry and the concept of limiting reactants.
Here's the step-by-step process to find the mass of NaCl:
1. Write down the balanced chemical equation for the reaction between AgNO3 and NaCl:
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
2. From the balanced equation, we can see that the molar ratio between AgNO3 and AgCl is 1:1. This means that for every 1 mole of AgNO3, we need 1 mole of AgCl.
3. Calculate the moles of AgNO3 present in the solution using the given concentration and volume:
Moles of AgNO3 = Concentration (M) × Volume (L)
Moles of AgNO3 = 0.100 M × 1.5 L = 0.15 moles AgNO3
4. Since the molar ratio between AgNO3 and AgCl is 1:1, we can say that we need 0.15 moles of AgCl for complete precipitation.
5. Using the molar mass of AgCl, calculate the mass of AgCl:
Molar mass of AgCl = Atomic mass of Ag + Atomic mass of Cl
= 107.87 g/mol + 35.45 g/mol
= 143.32 g/mol
Mass of AgCl = Moles of AgCl × Molar mass of AgCl
= 0.15 moles × 143.32 g/mol
≈ 21.50 g
Therefore, the mass of NaCl that must be added to 1.5L of 0.100M AgNO3 solution to precipitate all the Ag+ ions in the form of AgCl is approximately 21.50 grams.
AgNO3 + NaCl ==> AgCl + NaNO3
moles AgNO3 = M x L = ?
moles NaCl = moles AgNO3
grams NaCl = moles NaCl x molar mass NaCl