posted by Alex .
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?
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.
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.
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.
Thanks, I came up with 2308 mL
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.