2007 Practice Test #42,#43

[Delegate]

Hi all,

I'm having major trouble setting up stoich problems. They are from the 2007 practice exam. Could someone please point out the setup or basics for these?



42. If I mole of N2 and 1 mole of H2 are mixed and allowed to react according to the equation N2+3H2-->2NH3. What is the maximum number of moles of NH3 that could be produced?
a) 2/3
b)3/2
c)2/1
d)1/2
e)1/1

43. A flask weights 95g when empty. When filled with 200ml of a certain liquid, the weight is 328g. What volume (in ml) would 1000 g of the liquid occupy?



Thank you!

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[PooyaH]

Hi all,

I'm having major trouble setting up stoich problems. They are from the 2007 practice exam. Could someone please point out the setup or basics for these?



42. If I mole of N2 and 1 mole of H2 are mixed and allowed to react according to the equation N2+3H2-->2NH3. What is the maximum number of moles of NH3 that could be produced?
a) 2/3
b)3/2
c)2/1
d)1/2
e)1/1

43. A flask weights 95g when empty. When filled with 200ml of a certain liquid, the weight is 328g. What volume (in ml) would 1000 g of the liquid occupy?

44. If 3.00g of a nitrogen oxygen compound is found to contain 2.22g of oxygen, what is the percentage of nitrogen in the compound?

Thank you!

1. Since the reaction is showing 1 mole of N2 + 3 moles of H2 gives 2 moles of NH3, you need to divide H2 by 3 to get 1 mole of H2. Now, you also need to divide the moles of product (2NH3) by 3, so 1 mole on N2 and 1 mole of H2 should give you 2/3 moles of NH2.

2. You need to subtract 95 from 328 to get the weight of liquid first, so:
328 - 95 = 233g of liquid. Now since 233g = 200 ml, then 1000g = [(1000)(233)]/(200) = 1165ml

3. Since the compound weights 3.00g and contains 2.22g of oxygen, then 3.00 - 2.22 = 0.78g of nitrogen. And 0.78g / 3.00g = .26 so 26%.

 

[LetsGo2DSchool]

Oh boy, you need to do a lot of studying.

 

[PooyaH]

Oh boy, you need to do a lot of studying.

haha, don't give the kid a heart attack man. I used to be right where he is.

 

[ChrisM07]

You need to subtract 95 from 328 to get the weight of liquid first, so:
328 - 95 = 233g of liquid. Now since 233g = 200 ml, then 1000g = [(1000)(233)]/(200) = 1165ml

I solved it before I read your response, and we get different answers. I did:

328g-95g to give you 233g of liquid. (233g)/(200ml) gives you a density of 1.165g/ml.

Since we have the density, and a new given mass, when I solved for the volume, I get:

density=mass/volume

(1.165)=(1000)/(x)
x=858.38 ml.

What am I doing wrong? Maybe I need sleep

 

[PooyaH]

I solved it before I read your response, and we get different answers. I did:

328g-95g to give you 233g of liquid. (233g)/(200ml) gives you a density of 1.165g/ml.

Since we have the density, and a new given mass, when I solved for the volume, I get:

density=mass/volume

(1.165)=(1000)/(x)
x=858.38 ml.

What am I doing wrong? Maybe I need sleep

To be honest with you, I did think about that too, but the only thing that made me to go with the other reasoning was the calculation! If they have us do that kinda calculation on the DAT without a calculator, that's BS! Unless there's a big gap between the answer choices, so you could round the **** out of the numbers.

 

[Cofo]

This is how I solved the question too. I got 858.37
It's a basic density type of problem. Remember D = m/v
Then plug in the variables which they give you (the guy I quoted below...he did exactly what I did, so look at how he solved the problem)

I solved it before I read your response, and we get different answers. I did:

328g-95g to give you 233g of liquid. (233g)/(200ml) gives you a density of 1.165g/ml.

Since we have the density, and a new given mass, when I solved for the volume, I get:

density=mass/volume

(1.165)=(1000)/(x)
x=858.38 ml.

What am I doing wrong? Maybe I need sleep

 

[PooyaH]

This is how I solved the question too. I got 858.37
It's a basic density type of problem. Remember D = m/v
Then plug in the variables which they give you (the guy I quoted below...he did exactly what I did, so look at how he solved the problem)

I'm aware that it's a density related problem, but what I'm saying is since it was on the ADA 2007 DAT, I didn't think they would make us go through all that math. But yes, I think I should've stuck with the density reasoning.

 

[Dental2000]

1. Since the reaction is showing 1 mole of N2 + 3 moles of H2 gives 2 moles of NH3, you need to divide H2 by 3 to get 1 mole of H2. Now, you also need to divide the moles of product (2NH3) by 3, so 1 mole on N2 and 1 mole of H2 should give you 2/3 moles of NH2.

2. You need to subtract 95 from 328 to get the weight of liquid first, so:
328 - 95 = 233g of liquid. Now since 233g = 200 ml, then 1000g = [(1000)(233)]/(200) = 1165ml

3. Since the compound weights 3.00g and contains 2.22g of oxygen, then 3.00 - 2.22 = 0.78g of nitrogen. And 0.78g / 3.00g = .26 so 26%.

Im getting 858ml as well..

Heres a way to check if your doing the method right or not

Forget the numbers and just do your same calculation based on ur units

You've done (1000g x 233g)/200mL

which is ultimately giving g^2/mL which doesnt make sense at all
You need to get just mL which is simply 1000 * 200/233

 

[Dental2000]

Heres an easier method that i learnt called Unitary method

use this formulae

If a=b and c=d but you dont know what 'd' is?

then (a/c) = (b/d) =====> d = b*c/a

using plug and chug in our volume question

we know 233 g = 200 mL ( after doing the req subtraction)

we want to know the vol(say X) for 1000 g so

(233/1000) = (200/X) ====> X= 1000 * 200/233 = 858mL

Simply and easy...No need to find density