2 C4H10 + 13O2 > 8CO2 + 10H2O a) What is the mole ratio between butane and oxygen gas?
b) What is the mole ratio between water and oxygen gas?
c) How many moles of water formed?
d) How many moles of butane burned?
e) How many grams of butane burned?
f) How much oxygen was used up in moles?
g) How much oxygen was used up in grams?

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.

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