Answer:
1.99 grams of methane
Explanation:
A combustion reaction always produces the product carbon dioxide and water. So the chemical equation for this reaction would be:
CH₄ + O₂ → CO₂ + H₂O
To answer this question, you need to first determine which is the excess reactant. First step is to balance the equation:
CH₄ + 2O₂ → CO₂ + 2H₂O
Next we get the number of moles of each reactant we actually have:
8.00 g of Oâ‚‚ = ? moles of Oâ‚‚
You can get the number of moles, by first computing how many grams of the molecule is present in 1 mole.
O = 15.999g/mole. Since there are 2 oxygens in one mole of Oâ‚‚, all you need to do is add up the atomic mass of 2 oxygens. You will get:
Oâ‚‚= 31.998 g/mole
You can use this then to determine how many moles of Oâ‚‚ there are in 8.00g.
[tex]8.00g\times\dfrac{1mole}{31.998g}=0.250moles[/tex]
So there are 0.250 moles of Oâ‚‚ in 8.00 g of Oâ‚‚.
We do the same for CHâ‚„
4.00g of CHâ‚„=? moles of CHâ‚„
      C      H₄
CHâ‚„= 12.011 + 1.008(4) = 16.043 g/mole
[tex]4.00g\times\dfrac{1mole}{16.043g}=0.249moles[/tex]
So let's sum up our new given. We now have:
0.250 moles of Oâ‚‚
0.249 moles of CHâ‚„
Next we look at the molar ratio of reactants to produce products:
CH₄ + 2O₂ → CO₂ + 2H₂O
According to this equation we can assume the following:
We need 1 mole of CHâ‚„ for every 2 moles of Oâ‚‚ in this reaction. Using what we have, we will see how much of reactant of the other reactant we need to use up the other.
[tex]0.25molesofO_{2}\times\dfrac{1moleofCH_{4}}{2molesofO_{2}}=0.125molesofCH_{4}\\\\0.249molesofCH_{4}\times\dfrac{2molesofO_{2}}{1moleofCH_{4}}=0.498molesofO_{2}[/tex]
Compare the results with what we have:
What we have          What we need
0.250 moles of Oâ‚‚ Â < Â Â 0.498 moles of Oâ‚‚
0.249 moles of CHâ‚„ > Â Â 0.125 moles of CHâ‚„
This means that since we have less Oâ‚‚ that what we need to use up CHâ‚„, then Oâ‚‚ is the limiting reactant and CHâ‚„ is the excess.
To compute how much we have in excess, we use the number of moles produced when we use up limiting reactant which we did earlier and convert it into grams to determine how much in grams we used up.
Earlier we solved that we need 0.125 moles of CHâ‚„ to use up all the Oâ‚‚. Now convert that value into grams:
[tex]0.125molesofCH_{4}\times\dfrac{16.043gofCH_{4}}{1moleofCH_{4}}=2.005g of CH_{4}[/tex]
This means that 2.005g of CHâ‚„ will be used up.
Subtract that from the CHâ‚„ we already have:
4.00 g - 2.005 g =1.99 g of CHâ‚„
