17.8 g of CH4 react with 55.4 g of O2 to form CO2 and H20

identify the limiting and excess reagent. what is the mass of CO2 produced?
how much excess reactant is leftover?

Convert 17.8g CH4 to moles. moles = grams/molar mass.

Convert 55.4g O2 to moles.
Using the coefficients in the balanced equation, convert moles CH4 to moles CO2.
Do the same to convert moles O2 to moles CO2.
You will obtain two different answers; of course both can't be right. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
Then convert moles CO2 to grams. g = moles x molar mass.

How much of the excess reagent remains? Convert moles of the non-limiting reagent to moles of the excess reagent to determine the amount used then subtract from the original amount to determine the moles remaining, then convert to grams.

Post your work if you get stuck.

equation: CH4+ 2O2 = CO2 +2H20

1.11 moles of CH4
1.73 moles of O2

wait. why do we convert moles of CH4 to moles of CO2? and moles of O2 to moles of CO2?

So you can determine which produces the least CO2 and that will be the limiting reagent. There is a shorter way to do it but I don't like to use that method. Besides, you must determine grams CO2 produced anyway so it isn't much more trouble to do it this way.

okay, i was taught to pick one of the reactant and convert it to the other one.

so i got 1.11 mole of CO2 and 1.73 mole of CO2. so that means that CH4 is the limiting reagent?

1.11 mole of CO2 * 44.01 g of cO2 is 48.9 grams of CO2

wait, i think i did it wrong.

there is actually 0.865 moles of CO2 from the O2. SO the limiting reagent is O2?

CH4 + 2O2 ==> CO2 + 2H2O

Your second thought is right.
You have 1.11 moles CH4 which converts to 1.11 moles CO2.
You have 1.73 moles O2 which converts to 0.865 moles CO2; therefore, O2 is the limiting reagent. Then grams CO2 = 0.865 x 44.01 = ??

How much CH4 does 0.865 mole O2 use? That will be 0.865 x (1 mole CH4/2 moles O2) = 0.865 x (1/2)= about 0.45 (not exact) moles CH4. You had 1.11 to start so the difference is 1.11-0.45 = ? and that times 16 = grams CH4 unused.
Sorry I was late getting back to you; I was out of pocket about three hours.

oops. I didn't mean that for the last paragraph. How much CH4 is left?

You have 1.73 moles O2. That will use
1.73 moles O2 x (1 mole CH4/2 moles O2) = 1.73 moles O2 x (1/2) = 0.865 mols CH4 used. That leaves 1.11-0.865 = ? moles CH4 and that x molar mass CH4 = grams unreacted.