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March 28, 2017

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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 - ,

    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 - ,

    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 - ,

    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 - ,

    Thanks, I came up with 2308 mL

  • Chemistry - ,

    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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