Could some please tell me if I did this problem correctly?

Assume that at a given temperature, there is 1 g of CO2 gas in a one liter volume. How much mass of O2 would there be at the same temperature, in the same volume?

My answer:

Step 1)
Find the number of moles of carbon dioxide.

1mol/44.0095g
=. 022722mol/g

Because there is only 1g of carbon dioxide, there will .022722mol in the carbon monoxide in the sample.

Step 2)
Avogadro’s Hypothesis states that “ If the gas is an ideal gas, at a fixed temperature and pressure, a unit volume of the gas will have a fixed number of particles (a fixed number of moles in other words) regardless of what type of chemical it is.” This means that there will be the same number of moles of carbon monoxide as of oxygen. Therefore, there will be .022722mol of oxygen.

Step 3)
Convert moles of oxygen to grams.
(.022722mol)(31.9988g/mol)= .727077g

Answer: There will .727077g of oxygen at the same temperature and the same volume as 1g of carbon dioxide.

I didn't check your math, but the logic is exactly on target: same number of moles.

Thank you.

To verify if your answer is correct, let's go through the steps you took to solve the problem:

Step 1: Finding the number of moles of carbon dioxide.
You correctly used the molar mass of carbon dioxide (44.0095 g/mol) to find the number of moles in 1 gram of carbon dioxide. Your calculation of 0.022722 moles is correct.

Step 2: Applying Avogadro's hypothesis.
Avogadro's hypothesis states that equal volumes of gases, at the same temperature and pressure, contain equal numbers of particles (moles). Since we know the number of moles of carbon dioxide, we can conclude that there will also be 0.022722 moles of oxygen.

Step 3: Converting moles of oxygen to grams.
To find the mass of oxygen, you multiplied the number of moles (0.022722) by the molar mass of oxygen (31.9988 g/mol). Your calculation of 0.727077 grams is correct.

Based on your calculations, there will be 0.727077 grams of oxygen at the same temperature and volume as 1 gram of carbon dioxide. Therefore, your answer is correct.

Well done!