Posted by Kelsey on .
I have posted this once and still cannot get the correct answer.
Determine the mass of carbon monoxide that is produced when 45.6 of methane, CH4, react with 73.2 g of oxygen , O2. The products are carbon monoxide and water vapour.
This is a limiting reagent problem. You know that because BOTH reactants are given. Here is the system for working limiting reagent problems. Copy and memorize the steps.
1. Write a balanced equation.
2CH4 + 3O2 ==> 2CO + 4H2O
2a. convert 45.6 g CH4 to moles. moles = grams/molar mass.
45.6/16 = 2.85
2b. Do the same for moles O2. 73.2/32 = 2.2875
3a. Using the coefficients in the balanced equation, convert moles CH4 to moles CO.
2.85 mols CH4 x (2 moles CO/2 moles CH4) = 2 x (2/2) = 2x1= 2 moles CO.
3b. Do the same for moles O2.
2.2875 moles O2 x (2 moles CO/3 moles O2) = 2.2875 x (2/3) = 1.525 moles CO.
3c. Note that the answers for 3a and 3b don't give the same value for moles CO which means one of them is wrong. The correct answer in limiting reagent problems is ALWAYS the smaller value.
4. Convert the value from 3c to grams.
g = moles x molar mass. 1.525 x 28 = 42.70 g CO which I would round to 42.7. You need to confirm all of this as I estimated the molar masses of the above.