# How do you calculate the amount of substance from molar mass and volume [closed]

I am struggling with one calculation and it goes like this:

Calculate, how much of $$\ce{CO2}$$ is made during the reaction: $$\ce{2CO + O2}$$, if we use $$\pu{33,6 dm^3}$$ of $$\ce{O2}$$, also calculate the amount of $$\ce{CO}$$ used. M(CO)=28, M(CO2)=44.

The answer is $$\pu{132 g}$$ of $$\ce{CO2}$$ and $$\pu{67,2 L}$$. I especially don't know how I am supposed to get to the litres. Any help is appreciated :)

• A cubic decimeter is one liter. – Ed V May 25 at 18:47
• We do not do homework problems, or the like, here. You need to read the site’s homework policy and show your effort in solving the problem. We will help if that is done, but otherwise no. – Ed V May 25 at 18:52
• How do I ask homework questions on Chemistry Stack Exchange? – MaxW May 25 at 18:55
• Was there a temperature and pressure given for the gas? // As the problem is now presented you can solve for the volume of CO but not the grams. – MaxW May 25 at 19:00
• R3vive - Hint The oxygen gas and the carbon monoxide must be at the same temperature and pressure. Assume them both to be ideal gases. So for a stoichiometric reaction, what must the volume of carbon monoxide be? – MaxW May 25 at 19:48

The given question is a bad question.

Calculate, how much of $$\ce{CO2}$$ is made during the reaction: $$\ce{2CO + O2}$$, if we use $$\pu{33,6 dm^3}$$ of $$\ce{O2}$$, also calculate the amount of $$\ce{CO}$$ used. M(CO)=28, M(CO2)=44.

The OP gives the answer as 132 g of CO2, and 67,2 L of $$\ce{CO}$$.

The issues with the problems are:

• "...how much of $$\ce{CO2}$$ is made..."

For $$\ce{CO2}$$, this doesn't specify if it is the volume of gas, the number of moles, or the mass which is sought.

• "...also calculate the amount of $$\ce{CO}$$ used."

Again, this doesn't specify if it is the volume of gas, the number of moles, or the mass which is sought.

• The problem doesn't specify that $$\ce{CO}$$ and $$\ce{O2}$$ are present in stoichiometric proportions.

• The problem uses units $$\pu{dm^3}$$ for the $$\ce{O2}$$ but gives the answer for $$\ce{CO}$$ in $$\pu{Liters}$$.

• The problem doesn't specify that the reaction went to completion.

Utilizing only the information given, and the following assumptions, then the relative volumes of the gases could be found.

• That $$\ce{CO}$$ and $$\ce{O2}$$ are present in stoichiometric proportions.

• That the reaction went to completion.

• That after the reaction the gas volume was measured under the same conditions, pressure and temperature, used initially.

With the assumptions the volume of $$\ce{CO} = \pu{67.2 L}$$ and the volume of $$\ce{CO2} = \pu{67.2 L}$$.

It is to be noted that $$\pu{33.6 dm^3} = 1.5\times 22.4$$ where 22.4 is the old molar volume of an ideal gas at STP. So adding the assumption that the gases were measured using the old STP conditions allows the calculation of the grams of $$\ce{CO2}$$.

Don't know why but it seems that all the posters from India are still using the old STP specifications (before 1982).

Here is what I think is a better formulation of the question. It would assume that the solver understand that the old specification is used for STP, but that would be consistent throughout whatever book was being used.

Given that the reaction $$\ce{2CO + O2 -> 2CO2}$$ goes to completion, calculate the grams of $$\ce{CO2}$$ produced when, at STP, a stoichiometric amount of $$\ce{CO}$$ is added to $$\pu{33.6 dm^3}$$ of $$\ce{O2}$$. Also calculate the volume of $$\ce{CO}$$ added. M(CO)=28, M(CO2)=44.