What is the mass of the precipitate formed when 50 mL of 16.9\% solution of AgNO3 is mixed with 50 mL of 5.8\% NaCl solution ? (Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5)
- 3.5 g
- 7 g
- 14 g
- 28 g
Answer
(2) 7 g
Solution
AgNO3 + NaCl AgCl + NaNO3
Mass of AgNO3 in 50 mL = 16.9100 50 = 8.45 g
⇒ Moles of AgNO3 = 8.45169.8 = 0.0498 mol
Mass of NaCl in 50 mL = 5.8100 50 = 2.9 g
⇒ Moles of NaCl = 2.958.5 = 0.0496 mol
∴ NaCl is the limiting reagent, and moles of AgCl formed = 0.0496 mol
Molar mass of AgCl = 107.8 + 35.5 = 143.3 g mol-1
⇒ Mass of AgCl = 0.0496 143.3
= 7.1 g 7 g
Answer: (2)