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Chemistry 16 Online
OpenStudy (vera_ewing):

How many atoms of carbon are needed to produce 0.45 mol Al? 3C + 2Al2O3 -----> 4Al + 3CO2

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

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?

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

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 \(N_A\) \[N = nN_A\]

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

OpenStudy (vera_ewing):

thanks

OpenStudy (vera_ewing):

@UnkleRhaukus can you help with one more?

OpenStudy (unklerhaukus):

i can

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