Question

the reaction of aluminium chloride with silver acetate. what volume of 0.2M AlCl3 wpuld be needed to react completely with 40 ml of 0.5M AgOOCCH3 solution?

Answer 1

V ( AlCl₃ ) = ?

M ( AlCl₃) =  0.2 M

V ( AgC₂H₃O₂) = 40 mL in liters : 40 / 1000 => 0.04 L

M ( AgC₂H₃O₂ ) = 0.5 M

number of moles ( AgC₂H₃O₂ ) :

n = M x V

n = 0.5 x 0.04

n = 0.02 moles

Mole ratio :

3 AgC₂H₃O₂(aq) + AlCl₃(aq) = 3 AgCl(s) + Al(C₂H₃O₂)₃(aq)

3 moles ---------- 1 moles
0.02 moles ------ ? ( moles AlCl₃)

moles AlCl₃ = 0.02 x 1 / 3

moles AlCl₃ = 0.02 / 3

= 0.0066 moles

V( AlCl₃) =  n / M

0.2 = 0.0066 / V

V = 0.0066 / 0.2

V = 0.033 L or 33 mL

hope this helps!