Question

2 C4H10 + 13O2 -------> 8CO2 + 10H2O

Answer 1

Answer:

A. 2 : 13

B. 10 : 13

C. 10 moles of water, H₂O.

D. 2 moles of butane, C₄H₁₀

E. 116 g of butane, C₄H₁₀

F. 13 moles of oxygen, O₂

G. 416 g of oxygen, O₂

Explanation:

The equation for the reaction is given below:

2C₄H₁₀ + 13O₂ —> 8CO₂ + 10H₂O

A. Determination of the mole ratio between butane and oxygen gas.

Mole of butane, C₄H₁₀ = 2 moles

Mole of oxygen, O₂ = 13 moles

Mole ratio of butane and oxygen = 2 : 13

B. Determination of the mole ratio between water and oxygen gas

Mole of the water, H₂O = 10 moles

Mole of oxygen, O₂ = 13 moles

Mole ratio of water and oxygen = 10 : 13

C. Determination of the moles of water formed.

From the balanced equation above,

10 moles water, H₂O were produced.

D. Determination of the moles of butane burned.

From the balanced equation above,

2 moles of butane, C₄H₁₀ were burned.

E. Determination of the mass of butane burned.

Molar mass of C₄H₁₀ = (12×4) + (10×1)

= 48 + 10 = 58 g/mol

Mole of C₄H₁₀ = 2 moles

Mass of C₄H₁₀ =?

Mass = mole × molar mass

Mass of C₄H₁₀ = 2 × 58

Mass of C₄H₁₀ = 116 g

Thus, 116 g of butane, C₄H₁₀ were burned.

F. Determination of the number of mole of oxygen used.

From the balanced equation above,

13 moles of oxygen, O₂ were used.

G. Determination of the mass of oxygen used.

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mole of O₂ = 13 moles

Mass of O₂ =?

Mass = mole × molar mass

Mass of O₂ = 13 × 32

Mass of O₂ = 416 g

Thus, 416 g of oxygen, O₂ were used.