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.
