Question

5. Lead will react with hydrochloric acid as follows: + PbCl(s) + H2(g) 2 HCl(aq) + Pb(s) a. (10 points) Assign oxidation states to each atom, and identify the oxidizing agent and the reducing agent. b. (10 points) In the lab, a student mixes together 2.76 grams of lead with 1.12 grams hydrochloric acid. Determine the limiting reactant and the theoretical yield of solid lead (II) chloride (in grams).

Answer 1

A)

Oxidation state of H in reactant = +1

Oxidation state of Cl in reactant = -1

Oxidation state of Pb in reactant = 0

Oxidation state of H in product = 0

Oxidation state of Cl in product = -1

Oxidation state of Pb in product = +2

Oxidation state of H is decreasing.

So, H is reduced and HCl is oxidising agent.

Oxidation state of Pb is increasing.

So, Pb is oxidised and Pb is reducing agent.

Answer:

Oxidising agent: Pb

Reducing agent: HCl

B)

Molar mass of HCl,

MM = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass(HCl)= 2.76 g

use:

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(2.76 g)/(36.46 g/mol)

= 7.57*10^-2 mol

Molar mass of Pb = 207.2 g/mol

mass(Pb)= 1.12 g

use:

number of mol of Pb,

n = mass of Pb/molar mass of Pb

=(1.12 g)/(2.072*10^2 g/mol)

= 5.405*10^-3 mol

Balanced chemical equation is:

2 HCl + Pb ---> PbCl2 + H2

2 mol of HCl reacts with 1 mol of Pb

for 7.57*10^-2 mol of HCl, 3.785*10^-2 mol of Pb is required

But we have 5.405*10^-3 mol of Pb

so, Pb is limiting reagent

we will use Pb in further calculation

Molar mass of PbCl2,

MM = 1*MM(Pb) + 2*MM(Cl)

= 1*207.2 + 2*35.45

= 278.1 g/mol

According to balanced equation

mol of PbCl2 formed = (1/1)* moles of Pb

= (1/1)*5.405*10^-3

= 5.405*10^-3 mol

use:

mass of PbCl2 = number of mol * molar mass

= 5.405*10^-3*2.781*10^2

= 1.503 g

Answer:

Pb is limiting reagent

1.50 g