Chem Question

[thescreensavers]

Question 1: (ANSWERED)2.500 grams of MgSO4 ·[ ]xH2O, a hydrated salt with an unknown water content, is dried in an oven to remove the water. After drying, the anhydrous salt has a mass of 1.221 grams. How many moles of water are present per mole of hydrated magnesium sulfate?

The Answer is 7, but how do you get to that?


how do we figure out the water content? I found out that 1.279g of water was dried out when it was in the oven, after that I am lost. and I know how to get molecular weight. MgSO4 (weight 120g) H20 ( weight 18g) but do we add them? or what ?

Thanks

Edit: I got another question that's bugging me

Q #2 - In the lab, acid spills are often neutralized by adding sodium bicarbonate, What mass of Sodium Bicarbonate reacts with 225ml of 6.00 M HCL

3H+(aq)+3NaHCO3(s) ---> 3H20 + 3CO2(g)+3Na+(aq)

1:1 ratio, if I balanced it correctly 😛

HCL: 6.00M=X/.0255ml X = .135mol

then I did .135mol* 84(Mass of Sodium Bicarbonate) and I got 11.34

the answer is 113. But I cant seem to figure out the correct way to do it since that might just be a coincidence that I got a similar number.


Thanks agian

 

[sao123]

How many moles of water are present per mole of hydrated magnesium sulfate?

edit:

i didnt read the question properly...

use grams per mole to figure out the number of mols of water...
use grams per mole to figure out the number of mols of MsSO4
divide the two...

 

[frostedflakes]

Convert both quantities to moles. 1.221g MgSO4 is 0.0102 mol and 1.279g H2O is 0.0711 mol. Then divide moles of water by moles of magnesium sulfate to find the ratio (comes to like 6.97 for me).

 

[FelixDeCat]

Mole!!

 

[thescreensavers]

Bump for Question #2

 

[newb111]

Originally posted by: thescreensavers

Q #2 - In the lab, acid spills are often neutralized by adding sodium bicarbonate, What mass of Sodium Bicarbonate reacts with 225ml of 6.00 M HCL

3H+(aq)+3NaHCO3(s) ---> 3H20 + 3CO2(g)+3Na+(aq)

1:1 ratio, if I balanced it correctly 😛

HCL: 6.00M=X/.0255ml X = .135mol

then I did .135mol* 84(Mass of Sodium Bicarbonate) and I got 11.34

the answer is 113. But I cant seem to figure out the correct way to do it since that might just be a coincidence that I got a similar number.


Thanks agian

225ml = .225 L not .0255

 

[thescreensavers]

yup I just got it lol, damn it I just spend 30 min on that 🙁 fml ! only 3 more hours till the test.

Thanks for the help

 

[Possessed Freak]

For whatever reason, I just pictured my High School chem teacher saying: It's all stoichiometry. Let me stand on one foot and say that again... it's all stoichiometry.

 

[JTalbain]

Originally posted by: thescreensavers
Edit: I got another question that's bugging me

Q #2 - In the lab, acid spills are often neutralized by adding sodium bicarbonate, What mass of Sodium Bicarbonate reacts with 225ml of 6.00 M HCL

3H+(aq)+3NaHCO3(s) ---> 3H20 + 3CO2(g)+3Na+(aq)

1:1 ratio, if I balanced it correctly 😛

HCL: 6.00M=X/.0255ml X = .135mol

then I did .135mol* 84(Mass of Sodium Bicarbonate) and I got 11.34

the answer is 113. But I cant seem to figure out the correct way to do it since that might just be a coincidence that I got a similar number.

Your math is wrong at the HCl step. 225 mL is 0.225 L, not 0.0225 L. Correct that, and you get 1.35 mol of HCl, which is 113 g of NaHCO3.

 

[uclaLabrat]

Originally posted by: JTalbain

Originally posted by: thescreensavers
Edit: I got another question that's bugging me

Q #2 - In the lab, acid spills are often neutralized by adding sodium bicarbonate, What mass of Sodium Bicarbonate reacts with 225ml of 6.00 M HCL

3H+(aq)+3NaHCO3(s) ---> 3H20 + 3CO2(g)+3Na+(aq)

1:1 ratio, if I balanced it correctly 😛

HCL: 6.00M=X/.0255ml X = .135mol

then I did .135mol* 84(Mass of Sodium Bicarbonate) and I got 11.34

the answer is 113. But I cant seem to figure out the correct way to do it since that might just be a coincidence that I got a similar number.

Your math is wrong at the HCl step. 225 mL is 0.225 L, not 0.0225 L. Correct that, and you get 1.35 mol of HCl, which is 113 g of NaHCO3.

I taught you well, JT...all that work wasn't wasted! 😀

 

[BoomerD]

All this is exactly why I never took any kemistree classes. :roll: