Question

Butane reacts with oxygen gas to form carbon dioxide and water vapor as shown below:
2C₂H₁o + 130₂-8CO₂ + 10H₂O

a. If 20 g of butane were used in the experiment how many litres of water would be produced at SATP?

b. How many litres of oxygen would be consumed in this experiment?

c. If conditions were converted from SATP to STP, how many litres of water would now be produced?

Answer 1

A. The volume of water (in litres) produced at SATP is 42.73 L

B. The volume of oxygen consumed in the experiment is 55.55 L

C. If the conditions were converted from SATP to STP, the volume of water (in litres) produced is 38.64 L

How to determine the mole of C₂H₁₀

• Mass of C₂H₁₀ = 20 g
• Molar mass of C₂H₁₀ = (12×4) + (10×1) = 58 g/mol
• Mole of C₂H₁₀ = ?

Mole = mass / molar mass

Mole of C₂H₁₀ = 20 / 58

Mole of C₂H₁₀ = 0.3448 mole

A. How to determine the volume of water (in litres) produced at SATP

Balanced equation

2C₄H₁₀ + 13O₂ -> 8CO₂ + 10H₂O

From the balanced equation above,

2 moles of C₄H₁₀ reacted to produce 10 moles of water.

Therefore,

0.3448 mole of C₄H₁₀ will react to produce = (0.3448 × 10) / 2 = 1.724 mole of water

Thus, the volume of water (in litres) produced at SATP can be obtained by using the ideal gas equation as follow:

• Number of mole of water (n) = 1.724 mole
• Pressure (P) = SATP = 0.987 atm
• Temperature (T) = SATP = 298 K
• Gas constant (R) = 0.0821 atm.L/Kmol
• Volume of water produced (V) =?

PV = nRT

Divide both sides by P

V = nRT / P

V = (1.724 × 0.0821 × 298) / 0.987

Volume of water produced at SATP = 42.73 L

B. How to determine the volume of oxygen consumed

Balanced equation

2C₄H₁₀ + 13O₂ -> 8CO₂ + 10H₂O

From the balanced equation above,

2 moles of C₄H₁₀ reacted with 13 moles of O₂

Therefore,

0.3448 mole of C₄H₁₀ will react with = (0.3448 × 13) / 2 = 2.2412 moles of O₂

Thus, the volume of O₂ (in litres) consumed at SATP can be obtained by using the ideal gas equation as follow:

• Number of mole of O₂ (n) = 2.2412 moles
• Pressure (P) = SATP = 0.987 atm
• Temperature (T) = SATP = 298 K
• Gas constant (R) = 0.0821 atm.L/Kmol
• Volume of O₂ consumed (V) =?

PV = nRT

Divide both sides by P

V = nRT / P

V = (2.2412 × 0.0821 × 298) / 0.987

Volume of O₂ consumed = 55.55 L

C. How to determine the volume of water (in litres) produced at STP

• Number of mole of water (n) = 1.724 mole
• Pressure (P) = STP = 1 atm
• Temperature (T) = STP = 273 K
• Gas constant (R) = 0.0821 atm.L/Kmol
• Volume of water produced (V) =?

PV = nRT

Divide both sides by P

V = nRT / P

V = (1.724 × 0.0821 × 273) / 1

Volume of water produced at STP = 38.64 L

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