How many grams of silver chloride can be prepared by the reaction of 100.0 ml of 0.20 m silver nitrate with 100.0 ml of 0.15 m calcium chloride? calculate the concentrations of each ion remaining in solution after precipitation is complete.

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

a

explanation:

i'm assuming that the correct answer is indeed the first one/ "a"..