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May 14th, 2018, 02:05 AM   #1
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stoichiometry

equation 1: 2ZnS + 2O2 --> 2ZnO + SO2
equation 2: ZnO + CO --> Zn + CO2
300 tons of ZnS are processed each month by the plant. only 95% of the ZnO can be converted to Zn. Calculate the mass of Zinc produced in one month.


0.95 x 300 tons ZnS = 285 tons ZnS
n(ZnS) = m/M = 285,000,000g / 97.44g/mol = 2924877mol
Since ZnS:ZnO is 2:2, n(ZnO)=2924977mol
ZnO(eqn1):ZnO(eqn2)
2:1
Therefore, n(ZnO(eqn2))=1462438mol
ZnO:Zn
1:1
n(Zn)=1462438mol
m(Zn)= n x Mr = 95614224g = 95.6 tonnes

Is that correct?


part b) the plant burns hexene in an undersupply of oxygen to produce CO required:
Equation 3: C6H12 + 6O2 --> 6CO + 6H2O
What mass of hexene is required on a daily basis for the conversion of 10 tons of ZnO?


n(ZnO) = m/Mr = 100,000,00g / 81.38g/mol = 122880mol
ZnO:CO
1:1
n(CO) = 122880 mol
CO(eqn2):CO(eqn3)
1:6
n(CO(eqn3))= 737282 mol
C6H12:CO
1:6
n(C6H12) = 737282/6 = 122880mol
m(C6H12) = n x Mr = 122880 x 84.156 = 10341089g = 10.3 tonnes

I'm not sure if my solutions are correct.
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May 14th, 2018, 05:38 AM   #2
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Don't think I've ever seen such confusing variables and equations!
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May 14th, 2018, 06:16 AM   #3
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Quote:
Originally Posted by Denis View Post
Don't think I've ever seen such confusing variables and equations!
Oh, it's not so bad.

Here's a start:
First change 300 tons to metric (ie. grams). $\displaystyle 300 \text{ tons ZnS} = 2.72155 \times 10^{8} \text{ g ZnS}$.
So
$\displaystyle \frac{2.72155 \times 10^{8} \text{ g ZnS}}{1} \cdot \frac{1 \text{ mol ZnS}}{97.44 \text{ g ZnS}} \cdot \frac{2 \text{ mol ZnO}}{2 \text{ mol ZnS}} = 2.7931 \times 10^6 \text{ mol ZnO}$

So you can produce $\displaystyle 95 \% \cdot 2.7931 \times 10^6 \text{ mol ZnO} = 2.6954 \times 10^6 \text{ mol ZnO}$

The second step is done the same way. Can you finish?

-Dan

Last edited by skipjack; May 14th, 2018 at 07:49 AM.
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May 14th, 2018, 07:57 AM   #4
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Was 2ZnS + 3O$_2$ --> 2ZnO + 2SO$_2$ intended?
If "300 tons of ZnS" is correct, shouldn't the answer be given in tons rather than tonnes?

Was "10 tons of ZnO" intended to be "10 tonnes of ZnO"?
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May 14th, 2018, 10:22 AM   #5
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Quote:
Originally Posted by topsquark View Post
Oh, it's not so bad.

Here's a start:
First change 300 tons to metric (ie. grams). $\displaystyle 300 \text{ tons ZnS} = 2.72155 \times 10^{8} \text{ g ZnS}$.
So
$\displaystyle \frac{2.72155 \times 10^{8} \text{ g ZnS}}{1} \cdot \frac{1 \text{ mol ZnS}}{97.44 \text{ g ZnS}} \cdot \frac{2 \text{ mol ZnO}}{2 \text{ mol ZnS}} = 2.7931 \times 10^6 \text{ mol ZnO}$

So you can produce $\displaystyle 95 \% \cdot 2.7931 \times 10^6 \text{ mol ZnO} = 2.6954 \times 10^6 \text{ mol ZnO}$

The second step is done the same way. Can you finish?

-Dan
Why bother to change?

Why not just work in ton-molecules?
The unit of weight is irrelevant.
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May 14th, 2018, 10:38 AM   #6
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Originally Posted by studiot View Post
Why bother to change?

Why not just work in ton-molecules?
The unit of weight is irrelevant.
I'm a purist! (Okay, maybe not.) Anyway, I learned to do it in grams so I just think about it that way.

And yes, if the problem is given in tons then the answer should be also. Thanks for the correction.

-Dan
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