My Math Forum stoichiometry

 Chemistry Chemistry Forum

 May 14th, 2018, 03:05 AM #1 Member   Joined: May 2015 From: Australia Posts: 77 Thanks: 7 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.
 May 14th, 2018, 06:38 AM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,964 Thanks: 991 Don't think I've ever seen such confusing variables and equations!
May 14th, 2018, 07:16 AM   #3
Math Team

Joined: May 2013
From: The Astral plane

Posts: 2,041
Thanks: 815

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 Originally Posted by Denis Don't think I've ever seen such confusing variables and equations!

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 08:49 AM.

 May 14th, 2018, 08:57 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,288 Thanks: 1968 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"? Thanks from topsquark and studiot
May 14th, 2018, 11:22 AM   #5
Senior Member

Joined: Jun 2015
From: England

Posts: 891
Thanks: 269

Quote:
 Originally Posted by topsquark 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.

May 14th, 2018, 11:38 AM   #6
Math Team

Joined: May 2013
From: The Astral plane

Posts: 2,041
Thanks: 815

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 Originally Posted by studiot 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

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post pianist Chemistry 6 May 9th, 2018 05:33 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top