determine the mass of CO that is produced when 45.6g of CH4 react with 73.2g of O2.
2CH4 + 3O2 ---->2CO + 4H2O
nCH4 = 45.6 g/16.05 g/mol
=2.84 mol
nCO=2.84 mol
mCO=n*M
=2.84*28.01
=79.5 grams
I agree with your answer.
Please check this answer closely. Unless I have goofed, and I have been known to do that, this is a limiting reagent problem and you have not determined the limiting reagent. I think O2 is the limiting reagent; thus, your answer is too high by a factor of almost 2.
I failed to check for limiting reagent, and agree with DrBob
raspberries, snapping off on me. Years later, they had is still I'd surprise more than knew
to our to ramble off work It is let it go. his
To determine the mass of CO that is produced when 45.6g of CH4 react with 73.2g of O2, you need to follow these steps:
1. Determine the number of moles of CH4:
- The molar mass of CH4 (methane) is 16.05 g/mol.
- Divide the given mass of CH4 (45.6g) by its molar mass to find the number of moles.
- nCH4 = 45.6g / 16.05 g/mol = 2.84 mol
2. Use the balanced chemical equation to determine the stoichiometry between CH4 and CO:
- According to the balanced chemical equation: 2CH4 + 3O2 -> 2CO + 4H2O
- From the equation, you can see that 2 moles of CH4 produce 2 moles of CO.
- Therefore, the number of moles of CO produced (nCO) will be the same as the number of moles of CH4.
3. Calculate the mass of CO produced:
- Use the equation: mass = n * M
- nCO = 2.84 mol (from step 1)
- The molar mass of CO is 28.01 g/mol.
- mCO = nCO * M = 2.84 mol * 28.01 g/mol = 79.5 grams
Therefore, when 45.6g of CH4 reacts with 73.2g of O2, the mass of CO produced is 79.5 grams.