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Topic: Help please  (Read 15011 times)

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Offline bigdaddy

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Re: Help please
« Reply #15 on: August 31, 2013, 02:50:18 PM »
@Magician4 thanks so much i get it now!

Offline bigdaddy

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Re: Help please
« Reply #16 on: August 31, 2013, 03:45:12 PM »
Commercial brass, an alloy of Zn and Cu, reacts with hydrochloric acid as follows.
Zn(s) + 2 HCl(aq) --> ZnCl2(aq) + H2(g)
(Cu does not react with HCl.) When 0.5029 g of a certain brass alloy is reacted with excess HCl, 0.0901 g ZnCl2 is eventually isolated. and then i need the %Zn and %Cu

So i start by taking 0.0901g/136.2884g(1mol ZnCl2) = 0.00661098083182 mol, so 0.5029g is 0.00661098083182 mol of this Zn and Cu alloy, so the atomic weight of Zn is 65.382 so we take 0.00661098083182*65.382 = 0.43224g... so 0.43324/0.5029 gives me the percent of Zn to the total compound which is 86.15% and the % Cu would just be 100-86.15= 13.85 but apparently thats wrong.. can someone tell me where i went wrong?

Offline bigdaddy

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Re: Help please
« Reply #17 on: August 31, 2013, 04:04:43 PM »
Nitric acid is produced commercially by the Ostwald process, represented by the following equations.
4 NH3(g) + 5 O2(g) --> 4 NO(g) + 6 H2O(g)
2 NO(g) + O2(g) --> 2 NO2(g)
3 NO2(g) + H2O(l)  --> 2 HNO3(aq) + NO(g)
What mass in kg of NH3 must be used to produce 5.0 *10^6 kg HNO3 by the Ostwald process, assuming 100% yield in each reaction?

So I start by taking 5.0*10^6*1000/63.01296(1mol HNO3)= 7.935*10^7 mols

then i take that and find 7.935*10^7*17.03056(1 mol mass of NH3) = 1.3514*10^9 * 2 (because 4 of the NH3/ 2 of the HNO3) = 2702749872 / 1000 (to put back into kg) = 2.7*10^6.... but apparently thats wrong... does someone mind explaining where i went wrong....

Offline sjb

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Re: Help please
« Reply #18 on: August 31, 2013, 04:41:34 PM »
Nitric acid is produced commercially by the Ostwald process, represented by the following equations.
4 NH3(g) + 5 O2(g) --> 4 NO(g) + 6 H2O(g)
2 NO(g) + O2(g) --> 2 NO2(g)
3 NO2(g) + H2O(l)  --> 2 HNO3(aq) + NO(g)
What mass in kg of NH3 must be used to produce 5.0 *10^6 kg HNO3 by the Ostwald process, assuming 100% yield in each reaction?

So I start by taking 5.0*10^6*1000/63.01296(1mol HNO3)= 7.935*10^7 mols

then i take that and find 7.935*10^7*17.03056(1 mol mass of NH3) = 1.3514*10^9 * 2 (because 4 of the NH3/ 2 of the HNO3) = 2702749872 / 1000 (to put back into kg) = 2.7*10^6.... but apparently thats wrong... does someone mind explaining where i went wrong....

Does all the nitrogen in the ammonia end up in the nitric acid?

Offline sjb

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Re: Help please
« Reply #19 on: August 31, 2013, 04:43:44 PM »
Commercial brass, an alloy of Zn and Cu, reacts with hydrochloric acid as follows.
Zn(s) + 2 HCl(aq) --> ZnCl2(aq) + H2(g)
(Cu does not react with HCl.) When 0.5029 g of a certain brass alloy is reacted with excess HCl, 0.0901 g ZnCl2 is eventually isolated. and then i need the %Zn and %Cu

So i start by taking 0.0901g/136.2884g(1mol ZnCl2) = 0.00661098083182 mol, so 0.5029g is 0.00661098083182 mol of this Zn and Cu alloy, so the atomic weight of Zn is 65.382 so we take 0.00661098083182*65.382 = 0.43224g... so 0.43324/0.5029 gives me the percent of Zn to the total compound which is 86.15% and the % Cu would just be 100-86.15= 13.85 but apparently thats wrong.. can someone tell me where i went wrong?

Decimal places? If you have 0.09g of ZnCl2 does it make sense that this is made from 0.432g Zn?

Offline bigdaddy

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Re: Help please
« Reply #20 on: August 31, 2013, 05:41:28 PM »
omg it was supposed to be 0.000661098083182 not 0.00661098082182, lol i thought something was fishy xD... thanks sjb i wouldnt of caught that! i got it right now too

Offline Borek

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Re: Help please
« Reply #21 on: August 31, 2013, 05:53:38 PM »
Stop typing all those digits - they don't matter, they just make spotting similar mistakes much more difficult. If you are not expected to use significant digits (in which case you will need to determine number of digits you are going to type using SD rules), write just three or four (apart from leading zeros). It really makes life easier.
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Offline bigdaddy

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Re: Help please
« Reply #22 on: August 31, 2013, 06:10:30 PM »
@Borek xD alrighty i'm kinda ocd about exact values but ill start using SD

Offline Arkcon

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Re: Help please
« Reply #23 on: August 31, 2013, 07:26:20 PM »
@Borek xD alrighty i'm kinda ocd about exact values but ill start using SD
Yeah, but these values are silly.  You can't actually measure these many digits in real life -- no one makes a balance scale or glassware to those specs.  You'll have to cure yourself of this particular OCD, or you'll begin to lose points on exams, and you'll never be able to satisfy yourself with this many digits in a laboratory.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline Archer

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Re: Help please
« Reply #24 on: September 01, 2013, 02:55:46 AM »
@Borek xD alrighty i'm kinda ocd about exact values but ill start using SD

Your final answer cannot have more significant figures than the data you were provided with.

Here's a little exercise for you which I used to make my students do.

Calculate a problem using 1, 2, 3, 4, 5, etc significant figures. After a point the more you use the impact it has on the final answer becomes negligible.

Golden rule, if you're unsure use four
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Offline Arkcon

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Re: Help please
« Reply #25 on: September 01, 2013, 08:06:46 AM »
Golden rule, if you're unsure use four

A little archaic, but more apt than you might think.  The four decimal place rule dates back to when log tables and slide rules were used instead of hand-held calculators, and they were always to 4 decimal places.  Recently, I was using some published tables and noticed how many tables (as an example, the NBSB Alcoholometric Table) still use 4 decimal places.  I guess it just fits well on a page, and is accurate enough.  So bigdaddy:, you might want to focus your OCD onto 4 significant figures, and be more correct.
« Last Edit: September 10, 2013, 06:48:10 PM by Arkcon »
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

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