how many moles of C10H8 in 25g?
you will get 10 times that many moles of CO2
(c) you will use 1/12 mole of C10H8
b) Identify how many moles of carbon dioxide would be released from the equation in (a) if 25.0 g of naphthalene were burned in the presence of excess oxygen.
This is the equation: C10H8 + 12 O2 → 10 CO2 + 4 H2O
also can you help with the next question. Its related to the question above.
c) Explain what would happen if only 1 mole of oxygen gas were available to interact with naphthalene in this reaction.
4 answers
So all I have to do is find the the number of moles in CO2 and multiply it by ten?
and then for c) can explain a little more what you did their so I can understand better.
and then for c) can explain a little more what you did their so I can understand better.
the equation says that for every mole of C10H8 you get
10 moles of CO2
you use 12 moles of O2
10 moles of CO2
you use 12 moles of O2
To to ooblecks response:
C10H8 + 12 O2 → 10 CO2 + 4 H2O
If you know the number of moles in 25 g C10H8 the equation tells you that you get 10 mols CO2 for every 1 mol C10H8 burned.
For the question about using just 1 mol O2, the equation tells that every 1 mol C10H8 uses 12 mols O2; therefore, if you have just 1 mol O2 you can burn only 1/12 mol C10H8. I find that students often like to see the math so for the first one we have mols C10H8 = g/molar mass = 25/128 = 0.195, then mols CO2 = mols C10H8 x (10 mols CO2/1 mol C10H8) = 0.195 x 10/1 = ? Notice that mol C10H8 in the numberator cancel with mol C10H8 in the denominator to leave the answer in units of mols CO2 which is what you want. Applying this same reasoning to the second question you have
1 mol O2 x (1 mol C10H8/12 mol O2 = 1/12 mol C10H8. Notice that the units of O2 cancel to leave units of C10H8 which, again, is what you want for units in the answer.
C10H8 + 12 O2 → 10 CO2 + 4 H2O
If you know the number of moles in 25 g C10H8 the equation tells you that you get 10 mols CO2 for every 1 mol C10H8 burned.
For the question about using just 1 mol O2, the equation tells that every 1 mol C10H8 uses 12 mols O2; therefore, if you have just 1 mol O2 you can burn only 1/12 mol C10H8. I find that students often like to see the math so for the first one we have mols C10H8 = g/molar mass = 25/128 = 0.195, then mols CO2 = mols C10H8 x (10 mols CO2/1 mol C10H8) = 0.195 x 10/1 = ? Notice that mol C10H8 in the numberator cancel with mol C10H8 in the denominator to leave the answer in units of mols CO2 which is what you want. Applying this same reasoning to the second question you have
1 mol O2 x (1 mol C10H8/12 mol O2 = 1/12 mol C10H8. Notice that the units of O2 cancel to leave units of C10H8 which, again, is what you want for units in the answer.
1 answer