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Offline bigdaddy

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Help please
« on: August 30, 2013, 10:02:14 PM »
1)What number of atoms of phosphorus are present in 1.43 g of each of the compounds listed?
(angel) P4O6
(beer) Ca3(PO4)2
(coffee) Na2HPO4

2)Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly into the wound. If 23.0 g of Ag2O is reacted with 46.0 g of C10H10N4SO2, what mass of silver sulfadiazine, AgC10H9N4SO2, can be produced, assuming 100% yield?

3)Chloral hydrate (C2H3Cl3O2) is a drug formerly used as a sedative and hypnotic. It is the compound used to make "Mickey Finns" in detective stories.

What number of chlorine atoms are in 5.7 g chloral hydrate?

4)Hydrogen peroxide is used as a cleaning agent in the treatment of cuts and abrasions for several reasons. It is an oxidizing agent that can directly kill many microorganisms; it decomposes upon contact with blood, releasing elemental oxygen gas (which inhibits the growth of anaerobic microorganisms); and it foams upon contact with blood, which provides a cleansing action. In the laboratory, small quantities of hydrogen peroxide can be prepared by the action of an acid on an alkaline earth metal peroxide, such as barium peroxide.
BaO2(s) + 2 HCl(aq)  H2O2(aq) + BaCl2(aq)

What mass of hydrogen peroxide should result when 1.35 g of barium peroxide is treated with 25 mL of hydrochloric acid solution containing 0.0275 g of HCl per mL?

What mass of which reagent is left unreacted?

5)Consider the following unbalanced reaction.
P4(s) + F2(g)  PF3(g)
What mass of F2 is needed to produce 168 g of PF3 if the reaction has a 76.3% yield?

6)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.
(angel)

What is the composition of the brass by mass?

%Zn

%Cu


 How could this result be checked without changing the above procedure?

7)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(heart)   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?


please help im so lost on all of these

Offline Archer

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Re: Help please
« Reply #1 on: August 31, 2013, 01:29:04 AM »
You must at least have attempted the question(s), this is a Forum Rule.

 http://www.chemicalforums.com/index.php?topic=65859.0

See above for more details, you will also see that properly formatted formulae are preferred.

Here's a hint on how to start.

These questions are about atomic & molecular mass, moles and Avagadro's number. Read up on these three first.

Then start each problem by looking up the atomic weights of every atom mentioned in the question.
« Last Edit: August 31, 2013, 03:20:54 AM by Archer »
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Offline bigdaddy

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Re: Help please
« Reply #2 on: August 31, 2013, 07:16:20 AM »
i already did all of them and i know i did them wrong because i got wrong answers on all of these.... when i redid them i got the same answer, i just need someone to maybe do 1 and explain the stoichiometry and how it takes effect in the problem, i can probably get the rest after that

Offline bigdaddy

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Re: Help please
« Reply #3 on: August 31, 2013, 07:19:32 AM »
for example for #1 i would find the percent of phosphorous in relative to the total mass of 1 mol of the compound then multiply that by 1.43/1mol of P4O6, where did i go wrong?

Offline curiouscat

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Re: Help please
« Reply #4 on: August 31, 2013, 08:20:41 AM »
for example for #1 i would find the percent of phosphorous in relative to the total mass of 1 mol of the compound then multiply that by 1.43/1mol of P4O6, where did i go wrong?

That's vague. Give us your numbers.

Offline Archer

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Re: Help please
« Reply #5 on: August 31, 2013, 08:29:57 AM »
i already did all of them and i know i did them wrong because i got wrong answers on all of these.... when i redid them i got the same answer, i just need someone to maybe do 1 and explain the stoichiometry and how it takes effect in the problem, i can probably get the rest after that

That is not how this forum work I'm afraid.

If you want worked examples look in a text book.

If you want guidance from people on how to approach the problem the follow the Forum Rules and provide us with your workings and we will be able to highlight the errors which you have made.

This is the best way to learn.

If you got the answers wrong then we need to see how and where you made the mistakes let's start with question 1 and go from there...
“ I love him. He's hops. He's barley. He's protein. He's a meal. ”

Denis Leary.

Offline bigdaddy

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Re: Help please
« Reply #6 on: August 31, 2013, 09:26:37 AM »
for number 1 find number of atoms of phosphorous that are present in 1.43g of each of the compounds

P4O6
4P+6O=219.89163g/mol
then 1.43/219.89163=0.00650320341888

4P/P4O6

123.8950488g/219.8916g=0.563437

so (6.022*10^23)0.00650320341888*0.563437

which is 2.21*10^21…. but is wrong, please point out my mistake my thinking was that i take the ratio of the mass of the phosphorous to the total mass of the compound which is 56.3437% and find how many mols of the compound (P4O6) are in 1.46g so divide that by the the grams in 1 mol of P4O6 and then multiply both of those by 6.022*10^23


(if you can help me with this then i think i can get the other 2 in problem 1 and maybe a couple more i think my methodology may be incorrect but this is what i took from the book i'm using, however i'm afraid i may be far off.... sorry but this is my best attempt)
« Last Edit: August 31, 2013, 09:36:58 AM by bigdaddy »

Offline Borek

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Re: Help please
« Reply #7 on: August 31, 2013, 09:40:20 AM »
for number 1 find number of atoms of phosphorous that are present in 1.43g of each of the compounds

P4O6
4P+6O=219.89163g/mol
then 1.43/219.89163=0.00650320341888

You were OK up to this moment, but then you started to do strange things.

You know there are 0.0065 moles of P4O6.

How many P4O6 molecules is it?

How many atoms of P per each molecule of P4O6?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline bigdaddy

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Re: Help please
« Reply #8 on: August 31, 2013, 09:48:19 AM »
my next serious problem-Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly into the wound. If 23.0 g of Ag2O is reacted with 46.0 g of C10H10N4SO2, what mass of silver sulfadiazine, AgC10H9N4SO2, can be produced, assuming 100% yield?

So from what i understand i should take 23g/231.73587g(1mol Ag2O) = 0.09925092735967mols of Ag2O then i take 46g/250.2785(1mol C10H10N4SO2) = 0.18379525208917 now from my understanding my next step is to identify the limiting part which would be 0.09925092735967mol so then i take 0.09925092735967*357.1388(1molAgC10H9N4SO2) which is 35.45g which is apparently wrong.... someone please correct where im going wrong....

Offline curiouscat

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Re: Help please
« Reply #9 on: August 31, 2013, 09:50:45 AM »
my next serious problem-Silver sulfadiazine burn-treating cream creates a barrier against bacterial invasion and releases antimicrobial agents directly into the wound. If 23.0 g of Ag2O is reacted with 46.0 g of C10H10N4SO2, what mass of silver sulfadiazine, AgC10H9N4SO2, can be produced, assuming 100% yield?

So from what i understand i should take 23g/231.73587g(1mol Ag2O) = 0.09925092735967mols of Ag2O then i take 46g/250.2785(1mol C10H10N4SO2) = 0.18379525208917 now from my understanding my next step is to identify the limiting part which would be 0.09925092735967mol so then i take 0.09925092735967*357.1388(1molAgC10H9N4SO2) which is 35.45g which is apparently wrong.... someone please correct where im going wrong....

As an aside, try using a reasonable number of significant digits.

Offline bigdaddy

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Re: Help please
« Reply #10 on: August 31, 2013, 09:56:59 AM »

How many P4O6 molecules is it?

okay so would it be 0.0065*(6.022*10^23)=3.9143*10^21

How many atoms of P per each molecule of P4O6

so of the 10 atoms in the molecule would it be that 4/10 atoms of P per each molecule of P4O6 so i would take 4/10*3.9143*10^21 to get 1.56572*10^21?

Offline sjb

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Re: Help please
« Reply #11 on: August 31, 2013, 10:41:54 AM »

How many P4O6 molecules is it?

okay so would it be 0.0065*(6.022*10^23)=3.9143*10^21

How many atoms of P per each molecule of P4O6

so of the 10 atoms in the molecule would it be that 4/10 atoms of P per each molecule of P4O6 so i would take 4/10*3.9143*10^21 to get 1.56572*10^21?

Not quite. If you had 10 cakes, and each cake needed 4 oz of sugar and 6 oz of flour, how much sugar would you need?

Offline Borek

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Re: Help please
« Reply #12 on: August 31, 2013, 12:08:51 PM »
So from what i understand i should take 23g/231.73587g(1mol Ag2O) = 0.09925092735967mols of Ag2O then i take 46g/250.2785(1mol C10H10N4SO2) = 0.18379525208917 now from my understanding my next step is to identify the limiting part which would be 0.09925092735967mol so then i take 0.09925092735967*357.1388(1molAgC10H9N4SO2) which is 35.45g which is apparently wrong.... someone please correct where im going wrong....

You can't identify limiting reagent without knowing reaction equation (or at least molar ratio in which substances react).
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline magician4

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Re: Help please
« Reply #13 on: August 31, 2013, 02:04:22 PM »
your error beginns here:
Quote
So from what i understand i should take 23g/231.73587g(1mol Ag2O) = 0.09925092735967mols of Ag2O then i take 46g/250.2785(1mol C10H10N4SO2) = 0.18379525208917 now from my understanding my next step is to identify the limiting part which would be 0.09925092735967mol

... as one mole of Ag2O contains two silver atoms.
with each of your target molecules requiring only one of those, hence with every molecule of silverdioxide you could built two molecules of your target molecule

however, 2 * 0.099 > 0.183  :rarrow: you did identify the wrong limiting factor

additionally, even if your identification of this factor would have been correct (which it isn't), your subsequent calculation still would have been wrong:
Quote
so then i take 0.09925092735967*357.1388(1molAgC10H9N4SO2) which is 35.45g
as each mole Ag2O provides two Ag atoms, it should have been "2 * 0.09925092735967*357.1388 " instead


regards

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Offline bigdaddy

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Re: Help please
« Reply #14 on: August 31, 2013, 02:38:31 PM »
@sjb i got it now! your awesome thanks! xD its funny, i never thought about a molecule like a cake but now that i do everything seems easier

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