Question

A 30cm3 sample of butane, C4H10, was completely reacted in a limited supply of oxygen to produce 60 cm3 of carbon dioxide and 60 cm3 of carbon monoxide.
All volumes were measured at room temperature and pressure.
Which volume of oxygen was used?

A 90cm3 B 120cm3 C 150cm3 D 165cm3

(answer D, but why?)

(not sure what type of chemistry this is)

Answer 1

Answer on Question #69907 – Chemistry – Organic Chemistry

Question:

A 30cm³ sample of butane, C₄H₁₀, was completely reacted in a limited supply of oxygen to produce 60 cm³ of carbon dioxide and 60 cm³ of carbon monoxide.

All volumes were measured at room temperature and pressure.

Which volume of oxygen was used?

Solution:

Unbalanced equation of reaction is as follows:

$$C _ {4} H _ {1 0} + O _ {2} \rightarrow C O + C O _ {2} + H _ {2} O.$$

Let's balance it taking into account the fact that volumes of produced gases, carbon dioxide and carbon monoxide, are equal and thus their quantities (in numbers of molecules) are equal:

$$C _ {4} H _ {1 0} + n O _ {2} \rightarrow 2 C O + 2 C O _ {2} + 5 H _ {2} O$$

(Here numbers 2 preceding $CO$ and $CO_{2}$ are chosen to balance Carbon, and number 5 preceding $H_{2}O$ is chosen to balance Hydrogen.)

The unknown $n$ satisfies $2n = 2 + 2 \times 2 + 5$, hence $n = \frac{11}{2}$. Thus,

$$C _ {4} H _ {1 0} + \frac {1 1}{2} O _ {2} \rightarrow 2 C O + 2 C O _ {2} + 5 H _ {2} O$$

The volume of oxygen used is $\frac{11}{4}$ greater than volume of $CO$ or $CO_{2}$ produced:

$$V = \frac {1 1}{2} \times \frac {6 0 c m ^ {3}}{2} = 1 6 5 c m ^ {3}.$$

Answer:

165cm³

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