Question

13. The combustion of a sample of butane, C4H10 (lighter fluid), produced 2.76 grams of water.2 C4H10+ 1902 →The above give the reactants of a combustion reaction.a. What are the products?b.Balance the equation?c. If you start with 9.3 grams of C4H10, how much pure O₂ reacted?

Answer 1

Answer;

a) CO2 and H2O

c) 33.28 grams

Explanations

The combustion of an hydrocarbon will product carbondioxide and water as products.

a) The products are carbon dioxide(CO2) and water (H2O).

b) The balanced form of the reaction is given as shown below;

`2C_4H_(10)+13O_2\rightarrow8CO_2+10H_2O`

Since all the atoms of elements on both sides are equal, hence the equation is baanced.

c) Given the parameters

Mass of C4H10 = 9.3 grams

Calculate the moles of C4H10

moles of butane = mass/molar mass

moles of butane = 9.3/58.12

moles of butane = 0.16moles

According to stoichiometry, 2moles of butane reacts with 13moles of oxygen, the moles of oxygen required willl be;

`\begin{gathered} moles\text{ of O}_2=(13)/(2)*0.16 \\ moles\text{ of O}_2=1.04moles \end{gathered}`

Determine the mass of oxygen

Mass of oxygen = moles * molar mass

Mass of oxygen = 1.04 * 32

Mass of oxygen = 33.28grams

Hence the mass of oxygen that reacted is 33.28 grams