Answer:
Mass of AgCl = 2.87 g
[Ca²⁺] = 0.075 M
[Cl⁻] = 0.05 M
[NO₃⁻] = 0.10 M
Explanation:
The reaction between silver nitrate (AgNO₃) and calcium chloride (CaCl₂) is:
2AgNO₃(aq) + CaCl₂(aq) → 2AgCl(s) + Ca(NO₃)₂(aq)
The number of moles of the reactants are:
AgNO₃ = Volume (L)*concentration (M) = 0.1*0.2 = 0.02 mol
CaCl₂ = 0.1*0.15 = 0.015 mol
First, let's find which reactant is limiting. Testing for AgNO₃, the stoichiometry is:
2 moles of AgNO₃ ------------------------- 1 mol of CaCl₂
0.02 mol ------------------------- x
By a simple direct three rule:
2x = 0.02
x = 0.01 mol of CaCl₂, which is higher than was used, so CaCl₂ is in excess and AgNO₃ is the limiting reactant.
The stoichiometry between AgNO₃ and AgCl is:
2 moles of AgNO₃ ---------------- 2 moles of AgCl
0.02 mol ---------------- x
By a simple direct three rule
x = 0.02 mol of AgCl
The molar mass of AgCl is 143.32 g/mol, so the mass formed is:
m = 143.32*0.02 = 2.87 g
All the silver nitrate had reacted, and by the stoichiometry, only 0.01 mol of CaCl₂ reacted, so the number of moles remaining is:
nCaCl₂ = 0.015 - 0.01 = 0.005 mol
And the number of moles of Ca(NO₃)₂ formed is the same as CaCl₂ that reacted: 0.01 mol.
Both substances are in aqueous form, so, they produce ions:
CaCl₂ → Ca²⁺ + 2Cl⁻
By the stoichiometry: nCa²⁺ = 0.005 mol; nCl⁻ = 0.01 mol
Ca(NO₃)₂ → Ca²⁺ + 2NO₃⁻
By the stoichiometry: nCa²⁺ = 0.01 mol; nNO₃⁻ = 0.02 mol
The total volume is 0.2 L, so the ions concentrantions are:
[Ca²⁺] = (0.005 + 0.01)/0.2 = 0.075 M
[Cl⁻] = 0.01/0.2 = 0.05 M
[NO₃⁻] = 0.02/0.2 = 0.10 M