- Aug 3, 2005
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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
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