Answered

The combustion of a sample of butane, C4H10, produced 2.46 grams of water.
2C4H10 + 13O2 -> 8CO2 + 10H20

(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? And in grams?

Answer :

Answer:

(A) 0.14 moles water formed.

(B) 0.03 moles butane burned

(C) 5.7 grams of butane burned

(D) 0.18 moles of oxygen was used up in the reaction.

Explanation:

(A)

The given chemical reaction is as follows.

The number of water molecules in the reaction = 10

[tex]2C_{4}H_{10}+13O_{2}\rightarrow 8CO_{2}+10H_{2}O[/tex]

[tex]\frac{2.46g}{10(18)}=\frac{xH_{2}O}{10H_{2}O}= 0.14[/tex]

Therefore, 0.14 moles water formed.

(B)

Molarmass of butane = 180 g

The number of butane molecules in the reaction = 2

[tex]\frac{2.46g}{180}=\frac{xC_{4}H_{10}}{2C_{4}H_{10}}= 0.03[/tex]

Therefore, 0.03 moles butane burned

(C)

Molar mass of oxygen = 32g/mol

The number of water molecules in the reaction = 10

The number of oxygen molecules in the reaction = 13

[tex]\frac{2.46g}{10(18)}=\frac{xO_{2}}{13(32)}= 5.7g[/tex]

Therefore, 5.7 grams of butane burned

(D)

Molar mass of oxygen = 32 g/mol

Used amount of oxygen in the reaction = 5.7 g

[tex]\frac{5.7}{32}=0.18[/tex]

Therefore, 0.18 moles of oxygen was used up in moles.

mickymike92

Based on the data provided and the stoichiometry of the reaction;

  • moles of water produced is 0.136 moles.
  • moles of butane burned 0.027 moles
  • 4.900 grams of butane burned
  • moles of oxygen burbed is 0.177 moles
  • mass of oxygen burned is 5.66 g.

How can moles of a substance be determined?

The moles of a substance is determined using the formula:

  • moles = mass/molar mass

Molar mass of water = 18.0 g/mol

Mass of water = 2.46 g

Moles of water = 2.46/18

moles of water = 0.136 moles.

B. From the equation of the reaction, mole ratio of butane to wateris 2:10

Therefore;

moles of butane = 0.136 × 2/10

moles of butane = 0.027 moles

(C) Molar mass of butane = 180 g

mass of butane = 180 × 0.0272

Mass of butane = 4.900 g

Therefore, 4.900 grams of butane burned

(D)Molar mass of oxygen = 32 g/mol

From the equation of the reaction, mole ratio of oxygen and water = 13 : 10

moles of oxygen = 0.136 × 13/10

moles of oxygen = 0.177 moles

Mass of oxygen = 0.177 × 32

mass of oxygen = 5.66 g

Therefore, the amounts of reactants and products of the combustion of butane is derived from the stoichiometry of the reaction.

Learn more about moles and mass at: https://brainly.com/question/15374113

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