Question

I will give brainliest. If you burn the carbon in limited air, the reaction is

Answer 1

This reaction is different in that the carbon undergoes an incomplete combustion as opposed to complete combustion where carbon is fully oxidized. A caveat: incomplete combustion products in general can be difficult to predict without sufficient information, as it's not uncommon to obtain a mixture of different products.

Here, we are told that solid carbon is burned in limited air to produce a gas. I am presuming that, in the equation that's given, the "0" represents a blank where you must fill in a chemical symbol. In this case, our equation would be: 2C(s) + O₂(g) → 2CO(g).

There is not enough information here to provide the numerical answers to the two questions. From the words in the question (e.g., "is different" and "this time"), it would seem that this question is an excerpt from a larger or preceding question where specific numbers had been provided or computed.

However, it's possible to make some general observations on how one may go about answering these questions if one had more information.

Since we're to assume that oxygen is the limiting reagent, if one is given the amount of solid carbon (either in mass, moles, or number of atoms), it's possible to determine the moles of CO(g) that's produced since C and CO have an equal stoichiometric ratio. So, for example, if one burns 2 moles of C(s), then 2 moles of CO(g) would be produced.

But, there is still not enough information to compute the volume of CO gas if this is the line of questioning. We don't know, for instance, the temperature or pressure of the reaction conditions. In fact, the only way it would be possible to answer this would be if you were given beforehand a conversion factor that relates the volume of CO(g) to its quantity (e.g., to assume that one mole of gas occupies x liters).

As for the second question, this would depend on what you know about the quantity of the C(s) reacted and/or the quantity (or volume, from question a) of CO(g) produced. If you can get the number of moles of C(s) reacted or CO(g) produced, the number of moles of O₂(g) used up: It would be half the number of moles of C(s) reacted or half the number of moles of CO(g) produced). Again, it's impossible to determine the volume of O₂(g) using just the information provided here, so I suspect that you must have further information relating gas quantity to volume. As we did with CO(g), the volume of O₂(g) used up can be found using whatever conversion factor you have.

If you have any further information or questions, please feel free to follow up.