Posted by Alex on Thursday, September 6, 2007 at 6:46pm.
Mylanta is composed of aluminum hydroxide and calcium carbonate. assuming that a mylanta tablet contains 175 mg of aluminum hydroxide and 145 mg of cakcium carbonate, answer the following:
how many mL of 0.00836 M HCL (aq) can be neutralized with 2 Mylanta tablets.
I balanced the equation and multiplied the tablets by 2 to get 290 mg of CaCo3 and 350 mg Al(OH)3.
The I divided 290 by 200.02 g (is this the correct molecular weight of calcium carbonate. and multiplied this number by the mole ration of 2 moles HCl to 1 mole calcium carbonate to get 0.006 moles HCl, but now i'm unsure of what to do next?
Chemistry - DrBob222, Thursday, September 6, 2007 at 7:33pm
molar mass CaCO3 is closer to 100 than it is to 200. Redo that and recalculate.
mols CaCO3 x 2 is correct for mols HCl required to neutralize it.
Determine mols Al(OH)3. That will be 350/molar mass Al(OH)3.
mol HCl required to neutralize is mols Al(OH)3 x 3.
Add mols to neutralize HCl from CaCO3 to mols to neutralize HCl from Al(OH)3 to obtain total mols.
Then mols HCl = molarity HCl x volume HCl. Solve for volume HCl.
Post your work if you get stuck.
Chemistry - Anonymous, Thursday, September 6, 2007 at 8:37pm
Thank you. I redid the calculations and this is what i came up with.
i divided the .290 g of CaCO3 by 100.02 and multplied it by to to get 0.00580 moles HCl. Then I took .350 g of Al(OH)3 and divided it by 97.89 (was I supposed to multiply the aluminum by three? because i did but i'm not sure if that's the correct molecular weight) and multiplied that number by 3 to get .0107 moles HCl. Then I took the sum of .01107 moles and .00580 moles to get a total of .0165. After solving for the volume I got 1970 mL of HCl.
Chemistry - DrBob222, Thursday, September 6, 2007 at 8:46pm
0.00580 mols HCl for the CaCO3 is ok.
molar mass of Al(OH)3 can't be 98. Al is about 27 and OH is 17 so 3x17 = 51 and that plus 27 is about 78. So you need to redo that.
Yes, after finding mols Al(OH)3, multiply that by 3 to find mols HCl it will take to neutralize the Al(OH)3. Add them together and divide by 0.00836.
Chemistry - DrBob222, Thursday, September 6, 2007 at 9:07pm
I have 2304 but I may have used a slightly different value for molar masses. I used 100.08 for CaCO3 and 78 for Al(OH)3. Good problem.
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