Question

Aqueous solutions of iron(II) chloride and sodium sulfide react to form iron(II) sulfide and sodium chloride.

a. Write the balanced equation for the reaction.
b. If you combine 40. g each of Na2S and FeCl2, what is the limiting reactant?
c. What mass of FeS is produced?
d. What mass of Na2S or FeCl2 remains after the reaction?
e. What mass of FeCl2 is required to react completely with 40. g of Na2S?

Answer 1

#a. FeCl₂(aq) + Na₂S(aq) → FeS(s) + 2NaCl

(aq)

#b. FeCl₂ is the limiting reactant.

#c. 27.78 g

#d. 15.375 g Na₂S

#e. 65.023 g FeCl₂

Solution and explanation:

#a. Balanced chemical reaction

• An equation is balanced by putting coefficients on reactants and products to equate the number of atoms of each element is equal on both sides of the equation.
• The balanced chemical reaction between iron(II) chloride and sodium sulfide to form iron(II) sulfide and sodium chloride will be given by;

FeCl₂(aq) + Na₂S(aq) → FeS(s) + 2NaCl

(aq)

#b. The limiting reagent

We are given;

Mass of FeCl₂ = 40.0 g

Mass of Na₂S = 40.0 g

To determine the limiting reagent we first get the number of moles of each;

Step 1: Moles of FeCl₂

To calculate the number of moles we divide the mass of the compound by the molar mass.

Molar mass of FeCl₂ = 126.751 g/mol

Therefore;

Moles of FeCl₂ = 40 g ÷ 126.751 g/mol

= 0.316 moles

Step 2: Moles of Na₂S

Moles = Mass ÷ molar mass

Molar mass of Na₂S = 78.0452 g/mol

Therefore;

Moles of Na₂S = 40 g ÷ 78.0452 g/mol

= 0.513 moles

Therefore, the amount of FeCl₂ is way less than that of Na₂S, which means FeCl₂ is the limiting reactant.

#c. Mass of FeS

To calculate the mass of FeS we use the equation and mole ratio

From the equation;

1 mole of FeCl₂ reacts to produce 1 mole of FeS

Therefore;

0.316 moles of FeCl₂ would produce 0.316 moles FeS

But;

Mass = moles × Molar mass

Molar mass of FeS = 87.91 g/mol

Mass of FeS = 0.316 moles × 87.91 g/mol

= 27.78 g

Thus, the mass of FeS produced is 27.78 g

#d. Mass of Na₂S

Since FeCl₂ was the limiting reagent, then all of it was used in the reaction and therefore, no mass of FeCl₂ was left.

However, Na₂S was in excess

Moles of Na₂S that remained = 0.513 moles - 0.316 moles

= 0.197 moles

But, mass = moles × Molar mass

= 0.197 moles × 78.0452 g/mol

= 15.375 g

The mass of 15.375 g Na₂S remained.

#e. Mass of FeCl₂ required to react with 40 g of Na₂S

From the equation;

1 mole of FeCl₂ reacts with 1 mole Na₂S

Moles of Na₂S in 40 g Na₂S = 0.513 moles

Therefore, using the mole ratio;

Moles of FeCl₂ would be = 0.513 moles

But;

Mass = Moles × Molar mass

= 0.513 moles × 126.751 g/mol

= 65.023 g

Therefore, 40.0 g of Na₂S would completely react with 65.023 g FeCl₂