Question

The reaction of magnesium chloride with silver nitrate gives a precipitate of silver chloride.

MgCl2(aq) + 2AgNO3(aq) + Mg(NO3)2(aq) + 2AgCl(s)
A solution containing 0.001 mol of magnesium chloride reacts with excess silver nitrate.
What is the mass of the precipitate formed?
[Molar mass/gmol-': AgCl = 143.4]
A 0.0729
B 0.1439
C 0.287g
D 0.574g
(Total for Question 5 = 1 mark)

Answer 1

Mass = 0.287 g

Step-by-step explanation:

Given data:

Number of moles of MgCl₂ = 0.001 mol

Mass of precipitate formed (AgCl) = ?

Solution:

Chemical equation:

MgCl₂(aq) + 2AgNO₃(aq) → Mg(NO₃)₂(aq) + 2AgCl (s)

now we will compare the moles of MgCl₂ and AgCl.

MgCl₂ : AgCl

1 : 2

0.001 : 2/1×0.001 = 0.002

Mass of AgCl:

Mass = number of moles × molar mass

Mass = 0.002 mol × 143.4 g/mol

Mass = 0.287 g