Question

What is the theoretical yield (in g of precipitate) when 17.6 mL of a 0.501 M solution of iron(III) chloride is combined with 17.9 mL of a 0.614 M solution of lead(II) nitrate?

Nitrogen and oxygen react to form nitric oxide (NO). All substances are in the gas phase. If 0.448 atm of nitrogen and 0.371 atm of oxygen react, what is the partial pressure of nitric oxide (in mmHg) when this reaction goes 74.5 complete. The temperature and volume are constant.

Answer 1

1) Write the balanced chemical equation for the reaction of iron (III) chloride, FeCl₃ with lead (II) nitrate, Pb(NO₃)₂ as below.

2 FeCl₃ (aq) + 3 Pb(NO₃)₂ --------> 3 PbCl₂ (s) + 2 Fe(NO₃)₃ (aq)

As per the stoichiometric equation,

2 moles FeCl₃ = 3 moles Pb(NO₃)₂ = 3 moles PbCl₂.

Millimoles of FeCl₃ = (17.6 mL)*(0.501 M) = 8.8176 mmole.

Millimoles of Pb(NO₃)₂ = (17.9 mL)*(0.614 M) = 10.9906 mmole.

Find the limiting reactant as per the stoichiometry of the reaction.

Millimoles of Pb (NO₃)₂ that react with 8.8176 mmole FeCl₃: (8.8176 mmole FeCl₃)*(3 mole Pb(NO₃)₂/2 mole FeCl₃) = 13.2264 mmole.

Millimoles of FeCl₃ that react with 10.9906 mmole Pb(NO₃)₂: (10.9906 mmole Pb(NO₃)₂)*(2 mole FeCl₃/3 mole Pb(NO₃)₂) = 7.3271mmole.

Offcourse, we do not have 13.2264 mmole Pb(NO₃)₂ but we have more than 7.3271 mmole FeCl₃. Consequently, Pb(NO₃)₂ is the limiting reactant and the yield of the product is governed by the amount of the limiting reactant taken.

Theoretical amount of PbCl₂ in mmole = (10.9906 mmole Pb(NO₃)₂)*(3 mole PbCl₂/3 mole Pb(NO₃)₂) = 10.9906 mmole.

Molar mass of PbCl₂ = (1*207.2 + 2*35.453) g/mol = 278.106 g/mol.

Theoretical yield of PbCl₂ = (10.9906 mmole PbCl₂)*(1 mole/1000 mmole)*(278.106 g PbCl₂/1 mole PbCl₂) = 3.05655 g ≈ 3.0565 g (ans).