Question

14. 200 mL of 0.2 M NaCl solution is added to 200 mL of 0.5 M AgNO3 solution resulting in the formation of white precipitate. How many moles and grams of AgCl is formed? Also identify the limiting
reagent.​

Answer 1

Answer

n(AgCl)=0.04

w(AgCl) =5.72gm

Explanation:

NaCl+AgNO3—›AgCl+NaNO3

n(NaCl)=0.2×0.2=0.04

n(AgNO3)=0.2×0.5=0.1

Limiting reagent is NaCl

therefore,

n(AgCl)=0.04

w(AgCl)=0.04×143

=5.72gm

Answer 2

Answer:

0.04/5.72 g ; NaCl

Explanation:

NaCl + AgNO3 → AgCl + NaNO3

1 mol 1 mol 1 mol.

Molarity = moles / L

moles (NaCl) = 0.2 × 200/1000 = 0.04

moles (AgNo3) = 0.5 × 200/1000 = 0.10

since 0.04 < 0.10,

NaCl is the limiting reagent

now....

0.04 moles NaCl & 0.04 moles AgNo3 together form 0.04 moles of AgCl ( balanced equation above)

but...

moles (AgCl) = mass / molar mass

0.04 = mass / 143

mass of AgCl = 0.04 × 143

hence...

mass of AgCl = 5.72 g