Question

1. Magnesium chloride solution reacts with silver nitrate solution to form magnesium nitrate

solution and silver chloride.
Equation: MgCl2 (s) + 2 AgNO3 (aq) → Mg(NO3)2 (aq) + 2 AgCl (s)
Find the mass of silver chloride formed if
(a) 20 cm of 2.5 mol/dm^3 of magnesium chloride solution is used.

(6) 20 cm of 2.5 g/dm^3 of magnesium chloride solution is used.
​

Answer 1

a. 1,4332 g

b. 7.54~g

Further explanation

Given

Reaction

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

20 cm of 2.5 mol/dm^3 of MgCl2

20 cm of 2.5 g/dm^3 of MgCl2

Required

the mass of silver chloride - AgCl

Solution

a. mol MgCl2 :

`\tt 20~cm^3=20* 10^(-3)~dm^3\\\\mol=M* V\\\\mol=2.5~mol/dm^3* 20* 10^(-3)DM^3=0.05`

From equation, mol AgCl = 2 x mol MgCl2=2 x 0.05=0.1

mass AgCl(MW=143,32 g/mol)= 0.1 x 143,32=1,4332 g

b. mol MgCl2 (MW=95.211 /mol):

`\tt mol=M* V\\\\mol=(2.5~g/dm^3)/(95,211 g/mol)=0.0263~mol/dm^3`

From equation, mol AgCl = 2 x mol MgCl2=2 x 0.0263=0.0526

mass AgCl(MW=143,32 g/mol)= 0.0526 x 143,32=7.54~g