The following aqueous solutions of are mixed: 100.0 mL of 1.00 M lithium bromide and 100.0 mL of 1.00 M lead (II) nitrate. In an organized and clear manner, show all of your work and answers for this problem on the uploaded work. (a) Write the molecular equation for this reaction. (b) Write the total ionic chemical equation for this reaction. (c) Write the net ionic chemical equation for this reaction. (d) Identify the spectator ions in this reaction. (e) Identify the limiting reactant. (f) Determine the mass of solid product that is formed?

Answer and Explanation:

(a) When lithium bromide (LiBr) solution is mixed with a solution of lead (II) nitrate (Pb(NO₃)₂), lithium nitrate (LiNO₃) and lead (II) bromide (PbBr₂) are formed, according to the following molecular equation:

2LiBr(aq) + Pb(NO₃)₂(aq) → 2LiNO₃(aq) + PbBr₂(s) ↓

As the product PbBr₂ is an insoluble solid, it precipitates (↓).

(b) The total ionic equation is written with all ions of the reaction - no matter if they participate in the precipitate formation or not:

2Li⁺(aq) + 2Br⁻(aq) + Pb²⁺(aq) + 2NO₃⁻(aq) → 2Li⁺(aq) + 2NO₃⁻(aq) + PbBr₂(s)

(c) The net ionic equation is written including only the ions which participate in the precipitate formation. In this case, the precipitate is PbBr₂, and it is formed by Pb²⁺ and Br⁻ ions:

Pb²⁺(aq) + 2Br⁻(aq) → PbBr₂(s)

(d) The spectator ions are those which do not participate in the formation of the precipitate. From the total ionic equation, we can see that Li⁺ and NO₃⁻ ions are repeated on both sides of the equation, so they are redundant. Thus, the spectator ions are Li⁺ and NO₃⁻ ions.

(e) To identify the limiting reactant, we first calculate the moles of each compound as the product of the solution concentration and volume:

For LiBr:

C = 1.00 M = 1 mol/L

V = 100.0 mL x 1 L/1000 mL = 0.1 L

moles of LiBr = 0.1 L x 1.00 mol/L = 0.1 mol

The same for Pb(NO₃)₂:

C = 1.00 M = 1 mol/L

V = 100.0 mL x 1 L/1000 mL = 0.1 L

moles of Pb(NO₃)₂  = 0.1 L x 1.00 mol/L = 0.1 mol

From the total ionic equation, we can see that 2 mol of LiBr reacts with 1 mol of Pb(NO₃)₂ to give 1 mol of PbBr₂ (solid product). The stoichiometric molar ratio is 2 mol LiBr/1 mol Pb(NO₃)₂ and we have 0.1 mol of each reactant (0.1 mol LiBr/0.1 mol Pb(NO₃)₂= 1). As 2 mol LiBr/mol Pb(NO₃)₂ > 1 mol LiBr/mol Pb(NO₃)₂, LiBr is the limiting reactant.

(f) From the total ionic equation, we know that 2 moles of LiBr produce 1 mol of PbBr₂. To determine the mass of solid product (PbBr₂) formed, we first multiply the stoichiometric ratio (1 mol PbBr₂/2 mol LiBr) by the actual number of moles of limiting reactant we have (0.1 mol):

moles of PbBr₂ = 0.1 mol LiBr x (1 mol PbBr₂/2 mol LiBr) = 0.05 mol PbBr₂

Finally, we convert the moles of PbBr₂ to gram by using the molar mass of the compound:

Molar mass PbBr₂ = 207.2 g/mol + (2 x 79.9 g/mol) = 367 g/mol

grams of PbBr₂ = 0.05 mol x 367 g/mol = 18.35 g

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.