100cm 3 of ethyne was mixed with 240cm 3 of oxygen in a combustion chamber.

Explanation

To answer this question, we first need to write down the balanced chemical equation for the combustion of ethyne (C?H?) in oxygen (O?) to produce carbon(IV) oxide (CO?) and water (H?O):

\(2C_2H_2 + 5O_2 ? 4CO_2 + 2H_2O\)

Now, let's analyze the given information:

1. 100 cm\(^3\) of ethyne (C?H?)

2. 240 cm\(^3\) of oxygen (O?)

We need to find the volume of carbon(IV) oxide (CO?) produced.

From the balanced chemical equation, 2 moles of ethyne react with 5 moles of oxygen to produce 4 moles of carbon(IV) oxide. In this case, we have the volumes of ethyne and oxygen, so we need to determine which reactant is the limiting reactant.

First, let's convert the volumes to moles using the molar volume of a gas at standard conditions (22.4 L/mol or 22400 cm\(^3\)/mol):

Moles of ethyne = 100 cm\(^3\) / 22400 cm\(^3\)/mol = 0.00446 moles

Moles of oxygen = 240 cm\(^3\) / 22400 cm\(^3\)/mol = 0.0107 moles

Now, let's determine the limiting reactant:

Moles of ethyne needed = moles of oxygen