mass of oxygen produced from the combustion of 3X10^-3 of methane Solution What

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What mass of oxygen is needed for the complete combustion of 8.60×10-3 of methane? I'm guessing that you copy and pasted this from your "Mastering Chemistry" homework since they display units as .gifs. Step 1. Balance the equation using moles. Step 2. Stoichiometry Step 3. Rejoice, for you are correct! Let's see, What mass (g) of oxygen (O2, since it's one of the gases that exist as a pair when alone) is needed for the complete combustion of methane (CH4)? What do you always get from a combustion of most elements? (Carbon Dioxide and Water) O2 + CH4 --> CO2 + H2O The most important thing to do is to BALANCE the equation. (Some people create this cute little chart... try it without that way, it's too slow. Since O2 is alone, balance that last. Always start with non organic compounds if possible. [Things that are not O, C, H, etc...]) 1. C - already balanced O2 + CH4 --> CO2 + H2O 2. H - needs a 2 on the rightside O2 + CH4 --> CO2 + H2O O2 + CH4 --> CO2 + 2(H2O) 3. O - Now add a 2 to complete the balance (Notice I did the O last) O2 + CH4 --> CO2 + 2(H2O) 2(O2) + CH4 --> CO2 + 2(H2O) Balanced equation: 2(O2) + CH4 --> CO2 + 2(H2O) Step 2. Start by writing: Step a Step ß Step ? Step d 8.60×10-³ mol CH4 16.043 g CH4 2 mol O2 31.999 g O2 - - - - - - - - - - - - - x-------------------- x --------------- x--------------- 1 mol CH4 1 mol CH4 1 mol O2 Step a - Start with your units and write them in order so they cancel out diagonally. If you make sure you do this, you'll ace chemistry. Step ß - The number "16.043" is the amu (in the chart table) of CH4 written as g/mol. Since everything needs to cancel carry your units with the number! It's useful to know that "amu" is just "g/mol" and it's awesome that the top unit is the hint to the next part. Using this, you can cancel out that "mols CH4" Step ? - This is why you have to balance the equation. The mols will sometimes give you a ratio. What this step is saying is that for every 2mols of oxygen, you'll only have 1 mol of CH4 Step d - Now use O2's g/mol (called atomic mass or moler mass) and you'll end up with grams of O2. Also notice that I used 2 oxygen molecules because they don't exist alone. The answer: 8.830 g O2 (Round to the appropriate sigfig.) Hope that covers everything. Sorry if it seemed condecending, laying it out this much, but I didn't understand it until some did this for me. So now, I'm doing it for you.

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