Determine the mass of carbon dioxide that will be formed when 131.0 g of propane (C3H8) is reacted with 32 g of oxygen (O2)

C3H8 + 5O2 --> 3CO2 + 4H2O

I got O2 as my limiting reagent. Then I used the mol of that which is 1 mol to find the mol of CO2 which I got as 0.6 mol. My final answer is 26.4 g of CO2 but the answer sheet says something else. Am I right?

I think you can take your answer to the bank.

To determine the mass of carbon dioxide (CO2) formed, the first step is to identify the limiting reagent from the given reactants. In this case, you correctly identified oxygen (O2) as the limiting reagent.

To find the limiting reagent, you can compare the moles of the reactants to the stoichiometric ratio in the balanced chemical equation. The balanced equation indicates that 1 mole of propane (C3H8) reacts with 5 moles of oxygen (O2) to produce 3 moles of carbon dioxide (CO2).

Let's calculate the moles of oxygen available by using its given mass of 32 g. The molar mass of oxygen is 32 g/mol, so the number of moles is:

32 g O2 * (1 mol O2 / 32 g O2) = 1 mol O2

Since 1 mole of O2 is reacted in the balanced equation to produce 3 moles of CO2, we can determine the moles of CO2 produced. Thus:

1 mol O2 * (3 mol CO2 / 5 mol O2) = 0.6 mol CO2

At this point, your calculations are correct. You have determined that 0.6 moles of CO2 will be formed in the reaction.

Now, to obtain the mass of CO2 formed, you need to use the molar mass of carbon dioxide, which is 44 g/mol. Multiply the moles of CO2 by its molar mass:

0.6 mol CO2 * (44 g CO2 / 1 mol CO2) = 26.4 g CO2

Therefore, your answer of 26.4 g of CO2 is correct based on the calculations you provided.

If the answer sheet shows a different answer, please double-check the calculations or notify the instructor to clarify any discrepancies.