Question

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.

Answer 1

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

Answer 2

The concentrations of each ion remaining in solution after precipitation is complete is Mass of AgCl = 2.87 g, [Ca²⁺] = 0.075 M, [Cl⁻] = 0.05 m, [NO₃⁻] = 0.10 M.

What is silver chloride?

Silver chloride is an ionic compound in which the silver is cation and the chloride is anion.

The reaction is

$2AgNO_3(aq) + CaCl_2(aq) = 2AgCl(s) + Ca(NO_3)_2(aq)$

The number of moles of the reactants are:

$AgNO_3 = Volume (L) \times concentration (M) \\\\= 0.1 \times 0.2 = 0.02 mol$

$CaCl_2 = 0.1 \times 0.15 = 0.015 mol$

Now, finding the limiting reactant.

Testing for AgNO₃,

The stoichiometry is:

2 moles of AgNO₃  = 1 mol of CaCl₂

0.02 mol = x

By direct three rule:

2x = 0.02

$AgNO_3$ is the limiting reactant.

The stoichiometry between $AgNO_3$ and AgCl is:

2 moles of $AgNO_3$ =  2 moles of AgCl

0.02 mol = x

By a simple direct three rule

x = 0.02 mol of AgCl

The mass formed is

If the molar mass of AgCl is 143.32 g/mol

$m = 143.32\times0.02 = 2.87 g$

Remaining number of moles are

nCaCl₂ = 0.015 - 0.01 = 0.005 mol

The substances will produce ions, because they are aqueous form.

$CaCl_2 = Ca^2^++ 2Cl^-$

By the stoichiometry

nCa²⁺ = 0.005 mol

nCl⁻ = 0.01 mol

$Ca(NO_3)_2 = Ca^2^+ + 2NO_3^-$⁻

By the rule of stoichiometry

nCa²⁺ = 0.01 mol

nNO₃⁻ = 0.02 mol

The volume is 0.2 L,

The concentration of ions are

[Ca²⁺] =$\dfrac{(0.005 + 0.01)}{0.2} = 0.075 M$

[Cl⁻] = $\dfrac{0.01}{0.2} = 0.05 M$

[NO₃⁻] = $\dfrac{0.02}{0.2} = 0.10\; M$

Thus, the concentrations of each ion remaining in solution after precipitation is complete is Mass of AgCl = 2.87 g, [Ca²⁺] = 0.075 M, [Cl⁻] = 0.05 m, [NO₃⁻] = 0.10 M.

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