To determine the mass of Silver chloride
(AgCl) that gets precipitated, we need to first understand the reaction between Silver nitrate
(AgNO3) and Sodium chloride
(NaCl) to form Silver chloride and Sodium nitrate
(NaNO3).
The balanced chemical equation is:
AgNO3+NaCl⟶AgCl+NaNO3Given:
Volume of
AgNO3 solution
=150‌mlConcentration of
AgNO3 solution =
32%Volume of NaCl solution
=150‌mlConcentration of NaCl solution
=11%We will start by calculating the masses of both
AgNO3 and NaCl in their respective solutions.
Step 1: Calculate the mass of
AgNO3The mass percent concentration (w/v) gives the mass of the solute in grams per 100 mL of solution. Hence:
Mass of
AgNO3=(‌)×150‌ml=48gThe molar mass of
AgNO3 is calculated as:
108(Ag)+14(N)+16×3(O)=170g∕molThus, the number of moles of
AgNO3 is:
Moles of
AgNO3=‌=0.282‌mol Step 2: Calculate the mass of NaCl
Mass of
NaCl=(‌)×150‌ml=16.5gThe molar mass of NaCl is calculated as:
23(Na)+35.5(Cl)=58.5g∕molThus, the number of moles of NaCl is:
‌ Moles of ‌NaCl=‌=0.282‌mol Step 3: Determine the limiting reagent
Since both
AgNO3 and NaCl have the same number of moles ( 0.282 mol ) and they react in a 1:1 molar ratio, neither reagent is in excess. Therefore, both are completely consumed.
Step 4: Calculate the mass of AgCl precipitate
The number of moles of AgCl formed is the same as the moles of the limiting reagent, which is 0.282 mol .
The molar mass of AgCl is calculated as:
108(Ag)+35.5(Cl)=143.5g∕molThus, the mass of AgCl precipitated is:
Mass of
AgCl=0.282‌mol×143.5g∕mol=40.497g Conclusion:
Therefore, the mass of Silver chloride that gets precipitated is approximately 40.47 grams. Correct Answer: Option A - 40.47