Answer to Question #219379 in Chemistry for james

Question

Lead (II) nitrate reacts with sodium iodide react to produce a solid product lead (II) Iodide and sodium nitrate

(a) if 0.0830g of lead (II) nitrate solution is mixed with 0.300g sodium iodide solution, calculate the mass of the solid product made

(b) if 0.662 of lead (II) nitrate solution is mixed with 0.300g of sodium iodide solution, calculate the mass of the solid product made

Answer 1

The molar mass of Pb(NO₃)₂ is 331.2 g/mol

The molar mass of NaI is 149.89 g/mol

The molar mass of PbI₂ is 461.01 g/mol

Solution:

(a):

Calculate moles of each reactant:

(0.0830 g Pb(NO₃)₂) × (1 mol Pb(NO₃)₂ / 331.2 g Pb(NO₃)₂) = 0.00025 mol Pb(NO₃)₂

(0.300 g NaI) × (1 mol NaI / 149.89 g NaI) = 0.0020 mol NaI

Balanced chemical equation:

Pb(NO₃)₂(aq) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)

According to stoichiometry:

1 mol of Pb(NO₃)₂ reacts with 2 mol of NaI.

Thus, 0.00025 mol Pb(NO₃)₂ reacts with:

(0.00025 mol Pb(NO₃)₂ / 1 mol Pb(NO₃)₂) × (2 mol NaI) = 0.0005 mol NaI

However, initially there is 0.0020 mol of NaI (according to the task).

Therefore, Pb(NO₃)₂ acts as limiting reagent and NaI is excess reagent.

According to stoichiometry:

1 mol of Pb(NO₃)₂ produces 1 mol of PbI₂

Thus, 0.00025 mol Pb(NO₃)₂ produces:

(0.00025 mol Pb(NO₃)₂ / 1 mol Pb(NO₃)₂) × (1 mol PbI₂) = 0.00025 mol PbI₂

Calculate the mass of of the solid product (PbI₂):

(0.00025 mol PbI₂) × (461.01 g PbI₂ / 1 mol PbI₂) = 0.11525 g PbI₂ = 0.1153 g PbI₂

Answer (a): 0.1153 g PbI₂

(b):

Calculate moles of each reactant:

(0.662 g Pb(NO₃)₂) × (1 mol Pb(NO₃)₂ / 331.2 g Pb(NO₃)₂) = 0.0020 mol Pb(NO₃)₂

(0.300 g NaI) × (1 mol NaI / 149.89 g NaI) = 0.0020 mol NaI

Balanced chemical equation:

Pb(NO₃)₂(aq) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)

According to stoichiometry:

1 mol of Pb(NO₃)₂ reacts with 2 mol of NaI.

Thus, 0.0020 mol Pb(NO₃)₂ reacts with:

(0.0020 mol Pb(NO₃)₂ / 1 mol Pb(NO₃)₂) × (2 mol NaI) = 0.0040 mol NaI

However, initially there is 0.0020 mol of NaI (according to the task).

Therefore, NaI acts as limiting reagent and Pb(NO₃)₂ is excess reagent.

According to stoichiometry:

2 mol of NaI produces 1 mol of PbI₂

Thus, 0.0020 mol NaI produces:

(0.0020 mol NaI / 2 mol NaI) × (1 mol PbI₂) = 0.0010 mol PbI₂

Calculate the mass of of the solid product (PbI₂):

(0.0010 mol PbI₂) × (461.01 g PbI₂ / 1 mol PbI₂) = 0.46101 g PbI₂ = 0.4610 g PbI₂

Answer (b): 0.4610 g PbI₂