Answer to Question #320796 in Chemical Engineering for Janelle Idriz

Question

A barite composed of 100% BaSO4 is fused with carbon in the form of coke containing 6% ash (which is infusible). The composition of the fusion mass is

BaSO4 11.1%

BaS 72.8

C 13.9

Ash 2.2

_____

100%

Reaction:

BaSO4 + 4C → BaS + 4CO

Find the excess reactant, the percentage of the excess reactant, and the degree of completion of the reaction. 

Answer 1

Chemical reaction provided is as follows:

BaSO₄ + 4C → BaS + 4CO

This implies, 1 mol of BaSO₄ reacts with 4 mol of C to produce 1 mole of BaS and 4 mol of CO.

Consider that the initial weight of the substance is 100 g.

Calculate the weight of each substance from weight percentages as follows:

Weight percentage = weight / total weight x 100

For BaSO₄:

Weight percentage of BaSO₄ = weight of BaSO₄ / total weight x 100

11.1 = weight of BaSO₄ / 100 x 100

weight of BaSO₄ = 11.1 g

For BaS:

Weight percentage of BaS = weight of BaS / total weight x 100

72.8 = weight of BaS / 100 x 100

weight of BaS = 72.8 g

For C:

Weight percentage of C = weight of C / total weight x 100

13.9 = weight of C / 100 x 100

weight of C = 13.9 g

Now, calculate the moles of a substance as follows:

Moles of substance = mass of substance / molar mass

For BaSO₄:

Moles of BaSO₄ = weight of BaSO₄ / molar mass = 11.1 g / 233.38 g/mol = 0.047 mol

For BaS:

Moles of BaS = weight of BaS / molar mass = 72.8 g / 85.73 g/mol = 0.849 mol

For C:

Moles of C = weight of C / molar mass = 13.9 g / 12.0 g/mol = 1.158 mol

From the balanced chemical reaction provided above, 1 mol of BaSO₄ reacts with 4 mol of C to produce 1 mol of BaS and 4 mol of CO.

1 mol BaSO₄ = 1 mol BaS

So, 0.047 mol of BaSO₄ required to form 0.047 mol of BaS

Then, calculate the moles of C required to form 0.047 mol of BaS

1 mol of BaSO₄ = 4 mol of C

Moles of C = 0.047 mol of BaSO₄ x 4 mol of C / 1 mol of BaSO₄ = 0.188 mol of C

So, 0.047 mol of BaSO₄ reacted with 0.188 mol of C to produce 0.849 mol of BaS

Thus, the number of moles of C and BaSO₄ are 1.158 mol and 0.047 mol.

But, the required moles of C and BaSO₄ to form BaS are 0.047 mol of BaSO₄ and 0.188 mol of C.

Excess amount of C = 1.158 mol - 0.188 mol = 0.97 mol of C.

Thus, the limiting reactant is BaSO₄ and excess reactant is C.

The expression used to find the percentage of excess reagent is as follows:

Percent of excess reactant = (initial moles of C - used moles of C) / Initial moles of C x 100%

The initial moles of carbon are 1.158 mol and reacted moles of carbon are 0.188 mol of C.

Percent of excess reactant % = (1.158 mol - 0.188 mol) / 1.158 mol x 100% = 84.8 %

Hence, the percent of excess reactant is 84.8%

The expression used to find the value of degree of completion is as follows:

Degree of completion = (initial moles of reactants - used moles of reactants) / initial moles of reactants

Initial moles of reactants = 0.047 mol of BaSO₄ + 1.158 mol of C = 1.205 mol

Reacted moles = 0.047 mol of BaSO₄ + 0.97 mol of C = 1.02 mol

Degree of completion = (1.205 mol - 1.02 mol) / 1.205 mol = 0.360

Hence, the degree of completion is 0.360