The combustion of a sample of butane, C4H10 (lighter fluid), produced 2.46 grams of water. 2 C4H10 + 13O2 > 8CO2 + 10H2O
(a) How many moles of water formed?
(b) How many moles of butane burned?
(c) How many grams of butane burned?
(d) How much oxygen was used up in moles?
(e) How much oxygen was used up in grams?

0.137moles of water is formed, 0.0274 moles or 1.5892 g of butane burned and 0.1781 moles or 5.6992 g of O₂ is used.

Those reaction in which fuel is oxidized by the oxygen molecules and produce carbon dioxide and water molecule.

Given chemical reaction is:
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

Given mass of water = 2.46 grams

Moles will be calculated as:

n = W/M, where

W = given mass

M = molar mass

(a) Moles of water formed is calculated as:

Moles of water n = 2.46g / 18 g/mol = 0.137moles

(b) From the stoichiometry of the reaction, it is clear that:

10 moles of water = produced by 2 moles of butane

0.137 moles of water = produced by 2/10×0.137=0.0274 moles of butane

(c) Weight of butane is calculated by using moles:

W = 0.0274 × 58 g/mol = 1.5892 g

(d) From the stoichiometry of the reaction, it is clear that:

2 moles of butane = react with 13 moles of O₂

0.027 moles of butane = react with 13/2×0.027=0.1781 moles of O₂

(e) Mass of oxygen is calculated as:
W = 0.1781 x 32 g/mol = 5.6992 g

Hence, (a) 0.137moles, (b) 0.0274 moles, (c) 1.5892 g, (d) 0.1781 moles and (e) 5.6992 g.

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a. 0.137

b. 0.0274

c. 1.5892 g

d. 0.1781

e. 5.6992 g

Given

Reaction

2 C4H10 + 13O2 > 8CO2 + 10H2O

2.46 g of water

Required

moles and mass

Solution

a. moles of water :

2.46 g : 18 g/mol = 0.137

b. moles of butane :

= 2/10 x mol water

= 2/10 x 0.137

= 0.0274

c. mass of butane :

= 0.0274 x 58 g/mol

= 1.5892 g

d. moles of oxygen :

= 13/2 x mol butane

= 13/2 x 0.0274

= 0.1781

e. mass of oxygen :

= 0.1781 x 32 g/mol

= 5.6992 g