Question

When solutions of silver nitrate and calcium chloride are mixed, silver chloride precipitates out of solution according to the equation 2AgNO3(aq)+CaCl2(aq)→2AgCl(s)+Ca(NO3)2(aq) Part A What mass of silver chloride can be produced from 1.94 L of a 0.126 M solution of silver nitrate? Part B The reaction described in Part A required 3.49 L of calcium chloride. What is the concentration of this calcium chloride solution?

Answer 1

a)

moles of AgNO3 = M(AgNO3)*V(AgNO3)

= 0.126 M * 1.94 L

=0.244 mol

2AgNO3(aq)+CaCl2(aq)→2AgCl(s)+Ca(NO3)2(aq)

From balanced reaction above,

moles of AgCl formed = moles of AgNO3 reacted

= 0.244 mol

Molar mass of AgCl,

MM = 1*MM(Ag) + 1*MM(Cl)

= 1*107.9 + 1*35.45

= 143.35 g/mol

mass of AgCl,

m = number of mol * molar mass

= 0.244 mol * 143.35 g/mol

= 35.0 g

Answer: 35.0 g

b)

From balanced reaction,

moles of CaCl2 formed = (1/2)*moles of AgNO3

M(CaCl2)*V(CaCl2) = (1/2)*0.244 mol

M(CaCl2)*V(CaCl2) = 0.122

M * 3.49 L = 0.122

M = 0.0350 M

Answer: 0.0350 M