The approximate concentration of hydrochloric acid. HCL in the stomach (stomach acid) is 0.17M. Calculate the mass of the following antacids required to neutralize 50cm3 of the acid.
(a) bicarbonate of soda. NaHCO3
(b) Aluminum hydroxide Al (OH)3
Alternatively
Molar mass of NaHCO3 =23+1+12+(16*3)=84g/mol
84*0.17=14.28g
If 14.28g produces 1000cm³
Then xg will produce 50cm³
Cross multiply
14.28*50/1000=0.714g
Pls cross check
How many millimoles HCl do you have in 50 cc of the material?
That's mL x M = 0.17 x 50 = 8.5
NaHCO3 + HCl ==> NaCl + H2O + CO2
So you will need 8.5 millimoles NaHCO3 or 0.0085 mols.
grams NaHCO3 = mols x molar mass = ?
You do Al(OH)3 the same way but the equation is different.
Al(OH)3 + 3HCl ==> AlCl3 + 3H2O
I
I'll leave the remainder for you. Post your work if you get stuck.
8.5*84=714g
714/1000=0.714g
I hope I'm correct
Neither is correct.
First the NaHCO3. Here is what I told you about NaHCO3.
So you will need 8.5 millimoles NaHCO3 or 0.0085 mols.
grams NaHCO3 = mols x molar mass = ?
If I plug in 0.0085 x 84 I don't get anywhere near 63,513 grams. I doubt your local supermarket even stocks that much NaHCO3. Where did that 756.05949 come from? I don't see that you even used the 0.0085 mols.
Al(OH)3 is another story.
Grams NaHCO3 = mols X molar mass =
i.e 756.05949 X 84.0066 = 63,513.53gmol
For Al(OH)3= mols X molarmass
= 702.03202 X 78.01
= 54,765.52
pls help me cross-check
Don't mind me.
I actually converted 0.0085 to gram.
Thanks a lot for attending to my questions. may the Almighty bless abundantly.
I actually converted 0.0085 to gram.
You may have thought you converted mols to grams but you didn't I showed you how to do that. It is 0.0085 x 84 = ?
You get grams by mols x molar mass which is the 0.0085 x 84. That 756 is spurious number. Good luck to you in your classes and remember you can post here anytime. You may want to consider a tutor to help you get started..