Chemistry
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OpenStudy (vera_ewing):
How many atoms of carbon are needed to produce 0.45 mol Al?
3C + 2Al2O3 -----> 4Al + 3CO2
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OpenStudy (unklerhaukus):
your equation is balanced
3 moles of C make 4 moles of Al
OpenStudy (abhisar):
Hey there!
So from the equation we can see that 3 moles of carbon atoms are required to produce 4 moles of aluminium....ryt?
OpenStudy (unklerhaukus):
so, how many moles of C make one mole of Al?
OpenStudy (abhisar):
Use unitary method to find that...
OpenStudy (unklerhaukus):
3C = 4Al
Al = ?C
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OpenStudy (vera_ewing):
0.3375 moles of carbon?
OpenStudy (unklerhaukus):
yeah, that step is coming up, but , first we need to know how many mole of C make one mole of Al
OpenStudy (vera_ewing):
is it 2.6?
OpenStudy (unklerhaukus):
what?
OpenStudy (vera_ewing):
2.6 moles of C make one mole of Al?
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OpenStudy (unklerhaukus):
no, there should be less moles carbon than moles of aluminium
OpenStudy (vera_ewing):
i have no idea :(
OpenStudy (unklerhaukus):
3C = 4Al
Al = (3/4)C
OpenStudy (vera_ewing):
ok now what?
OpenStudy (unklerhaukus):
now
0.45 Al = ? C
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OpenStudy (vera_ewing):
0.3375 moles of carbon
OpenStudy (unklerhaukus):
yeah that's right
0.3375 moles of C make 0.45 moles of Al
OpenStudy (unklerhaukus):
so now you just need to convert 0.3375 moles into atoms,
use Avogadro's number \(N_A=6.022\times10^{23}[\text {molecules}/\text{mol}]\)
OpenStudy (vera_ewing):
ok so i got 2.6 × 10^26 atoms
OpenStudy (unklerhaukus):
The number \(N\) of atoms (molecules) of a substance is equal tot the number of moles \(n\), times Avogadro's number \(N_A\)
\[N = nN_A\]
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OpenStudy (vera_ewing):
i'm so confused. what did you get?
OpenStudy (unklerhaukus):
\[N_\text C = n_\text CN_A\\
\ \ \quad= 0.3375[\text{mol}]\times 6.022\times10^{23}[\text{molecules/mol}]
\]
OpenStudy (vera_ewing):
@UnkleRhaukus would the answer be 2.0 * 10^23 atoms
OpenStudy (vera_ewing):
@Abhisar ?
OpenStudy (unklerhaukus):
that's right
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OpenStudy (vera_ewing):
thanks
OpenStudy (vera_ewing):
@UnkleRhaukus can you help with one more?
OpenStudy (unklerhaukus):
i can