Answer to Question #175789 in Chemistry for Jeremy

Question

Lesson 16 1.Zinc reacts with hydrochloric acid based on the following equation:

Zn + 2HCl = ZnCl2 + H2

If 150 g of Zn reacts with 73 g of HCl:

a. Which reactant is the limiting reagent?

b. What mass of ZnCl₂ would be produced?

2.If 10.00 g of magnesium is reacted with 95.75 g of copper (II) sulphate, magnesium sulphate and copper are formed.

Mg + CuSO4 = MgSO4 + Cu

a. Which reactant is in excess?

b. Calculate the mass of copper formed.

3.Metallic silver can be prepared by the reaction of hydrogen gas with silver oxide.

Ag2O + H2 = 2Ag + H2O

Silver oxide is a very expensive reactant compared to hydrogen gas. If you were to carry out this reaction in the lab in the most cost effective way to produce a high yield, which of the reactants should be used in excess?

Answer 1

Solution:

(Question 1a):

The balanced chemical equation:

Zn + 2HCl → ZnCl₂ + H₂

The molar mass of Zn is 65.38 g mol^-1.

The molar mass of HCl is 36.458 g mol^-1.

Determine moles of each reactant:

(150 g Zn) × (1 mol Zn / 65.38 g Zn) = 2.2943 mol Zn

(73 g HCl) × (1 mol HCl / 36.458 g HCl) = 2.0023 mol HCl

According to the chemical equation above: n(Zn) = n(HCl)/2

Choose one reactant and determine how many moles of the other reactant are necessary to completely react with it. Let's choose HCl:

n(Zn) = n(HCl) / 2 = 2.0023 mol / 2 = 1.00115 mol

The calculation above means that we need 1.00115 mol of Zn to completely react with 2.0023 mol of HCl.

We have 2.2943 mol Zn and therefore more than enough zinc.

Thus zinc (Zn) is in excess and hydrochloric acid (HCl) must be the limiting reagent.

The limiting reagent is hydrochloric acid (HCl).

(Q1b):

The molar mass of ZnCl₂ is 136.3 g mol^-1.

Since hydrochloric acid (HCl) is a limiting reagent, then:

(73 g HCl) × (1 mol HCl / 36.458 g HCl) × (1 mol ZnCl₂ / 2 mol HCl) × (136.3 g ZnCl₂ / 1 mol ZnCl₂) = 136.457 g ZnCl₂ = 136.5 g ZnCl₂

136.5 grams of ZnCl₂ would be produced.

(Q2a):

The balanced chemical equation:

Mg + CuSO₄ → MgSO₄ + Cu

The molar mass of Mg is 24.305 g mol^-1.

The molar mass of CuSO₄ is 159.609 g mol^-1.

Determine moles of each reactant:

(10.00 g Mg) × (1 mol Mg / 24.305 g Mg) = 0.41 mol Mg

(95.75 g CuSO₄) × (1 mol CuSO₄ / 159.609 g CuSO₄) = 0.5999 mol CuSO₄ = 0.60 mol CuSO₄

According to the chemical equation above: n(Mg) = n(CuSO₄)

Choose one reactant and determine how many moles of the other reactant are necessary to completely react with it. Let's choose Mg:

n(CuSO₄) = n(Mg) = 0.41 mol

The calculation above means that we need 0.41 mol of CuSO₄ to completely react with 0.41 mol of Mg.

We have 0.60 mol CuSO₄ and therefore more than enough copper (II) sulphate.

Thus copper (II) sulphate (CuSO₄) is in excess and magnesium (Mg) must be the limiting reagent.

Copper (II) sulphate (CuSO₄) is in excess.

(Q2b):

The molar mass of Cu is 63.546 g mol^-1.

Since magnesium (Mg) is a limiting reagent, then:

(10.00 g Mg) × (1 mol Mg / 24.305 g Mg) × (1 mol Cu / 1 mol Mg) × (63.546 g Cu / 1 mol Cu) = 26.15 g

26.15 grams of copper (Cu) formed.

(Q3):

The balanced chemical equation:

Ag₂O + H₂ → 2Ag + H₂O

Hydrogen gas (H₂) should be used in excess.

Answers:

(Q1a): The limiting reagent is hydrochloric acid (HCl).

(Q1b): 136.5 grams of ZnCl₂ would be produced.

(Q2a): Copper (II) sulphate (CuSO₄) is in excess.

(Q2b): 26.15 grams of copper (Cu) formed.

(Q3): Hydrogen gas (H₂) should be used in excess.