Write a balanced chemical equation for each of the following. (A) Solid copper reacts with solid sulfur to form solid copper (I) sulfide. Express your answer as a chemical equation. Identify all of the phases in your answer.

(B) Liquid octane reacts with oxygen gas to form carbon dioxide gas and water vapor.

(C) Sulfur dioxide gas reacts with oxygen gas to form sulfur trioxide gas.

A) 2Cu (s) + S (s) → Cu2S (s)

B) 2C8H18 (l) + 25O2 (g) → 16CO2 (g) + 18H2O (g)

C) 2SO2 (g) + O2 (g) → 2SO3 (g)

Explanation:

(A) Solid copper reacts with solid sulfur to form solid copper (I) sulfide.

Step 1: Data given

Solid copper = Cu (s)

solid sulfur = S (s)

solid copper (I) sulfide = Cu2S

Step 2: balancing the equation

Cu (s) + S (s) → Cu2S (s)

On the left side we have 1x Cu, on the right side we have 2x Cu (in Cu2S). To balance the amount of Cu we have to multiply Cu on the left side by 2.

2Cu (s) + S (s) → Cu2S (s)

(B) Liquid octane reacts with oxygen gas to form carbon dioxide gas and water vapor.

Step 1: Data given

liquid octane = C8H18 (l)

oxygen gas = O2 (g)

carbon dioxide gas = CO2 (g)

water vapor = H2O (g)

Step 2: Balancing the equation

C8H18 (l) + O2 (g) → CO2 (g) + H2O (g)

On the left side we have 8x C (in C8H18), on the right side we have 1 x C (in CO2). To balance the amount of C on both sides, we have to multiply CO2 on the right side by 8.

C8H18 (l) + O2 (g) → 8CO2 (g) + H2O (g)

On the left side we have 36x H (in 2C8H18), on the right side we have 2x H (in H2O). To balance the amount of H on both sides, we have to multiply H2O on the right side by 18.

C8H18 (l) + O2 (g) → 8CO2 (g) + 9H2O (g)

On the left side we have 2x O (in O2), on the right side we have 25x O (16x in CO2 and 9x in 9H2O). To balance the amount of O on both sides, we have to multiply O2 on the left side by 25. And all other coeffcients by 2.

2C8H18 (l) + 25O2 (g) → 16CO2 (g) + 18H2O (g)

(C) Sulfur dioxide gas reacts with oxygen gas to form sulfur trioxide gas.

Step 1: Data given.

Sulfur dioxide gas = SO2 (g)

oxygen gas = O2 (g)

sulfur trioxide gas = SO3 (g)

Step 2: Balancing the equation

SO2 (g) + O2 (g) → SO3 (g)

On the left side we have 4x O (2x in SO2 and 2x in O2) and 3x O on the right side. To balance the amount of O on both sides, we have to multiply SO2 and SO3 both by 2.

2SO2 (g) + O2 (g) → 2SO3 (g)

A. 2Cu (s) + S (s) → Cu₂S (s)

B. 2C₈H₁₈ (g) + 25O₂ (g) → 16CO₂ (g) + 18H₂O (g)

C. 2SO₂ (g) + O₂ (g) → 2SO₃ (g)

Explanation:

A. Let's think the reactants:

Cu and S

The product is: CuS

Equation: 2Cu (s) + S (s) → Cu₂S (s)

B. The reactants are:

O₂ and C₈H₁₈

The products are: water and CO₂

This is a combustion reaction: 2C₈H₁₈ (g) + 25O₂ (g) → 16CO₂ (g) + 18H₂O (g)

C. Reactants: SO₂ and O₂

Products: SO₃

Equation: 2SO₂ (g) + O₂ (g) → 2SO₃ (g)