How many grams of CO2 are produced when 11.47 moles hexane combust?

My work:I balanced the equation like so C6H14 + 19/2 O2 ==> 6CO2 + 7H2O

My teacher taught us the 19/2 was valid

My work:

11.47 mol C6H14 * 6 CO2/1 mol C6H14 * 44g/mol/1 mol CO2 = 3, 028 g of CO2

I agree with your answer. As for the 19/2, I think many profs will not allow it but I don't see anything wrong with it.

Agreed. In high school I was taught something else. Like multiply every thing by 2. To get a whole number.

Thank you for your help

To find the number of grams of CO2 produced when 11.47 moles of hexane (C6H14) combusts, you need to use the balanced equation for the combustion reaction:

C6H14 + 19/2 O2 → 6 CO2 + 7 H2O

Based on the balanced equation, 1 mole of hexane (C6H14) produces 6 moles of carbon dioxide (CO2).

To start, convert moles of hexane to moles of carbon dioxide using a conversion factor:

11.47 mol C6H14 * 6 mol CO2 / 1 mol C6H14 = 68.82 mol CO2

Next, convert moles of carbon dioxide to grams of carbon dioxide using the molar mass of CO2:

68.82 mol CO2 * 44 g/mol = 3,028 g CO2

Therefore, when 11.47 moles of hexane combusts, approximately 3,028 grams of carbon dioxide are produced.