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Chemistry CHALLENGED

  1. Oct 13, 2004 #1
    HELP!!!!!

    I need to know how to figure these problems out, while showing all
    work so I can follow the steps....thanks for the help!!!!!!!

    1. how many grams of Mg(OH)2 (2 is a subscript) will be needed to neutralize
    25ml of stomach acid if stomach acid is 0.10 M HCl?

    2. How many ml of a 0.10 NaOH solution are needed to neutralize 15ml of 0.20
    M H3PO4 (subscripts) solution???
     
  2. jcsd
  3. Oct 13, 2004 #2
    Please show your work first.
     
  4. Oct 14, 2004 #3
    All i can say if please use:

    Ma x Mb / Va x Vb transpose for what you need.
     
  5. Oct 15, 2004 #4
    for number 1:

    find out how many moles of HCl are in 25mL of a 0.10M solution by multiplying volume times molarity.

    M HCl = 25mL x 1L/1000mL x 0.10mol / 1L

    next calculate how many mL of xM [tex]Mg(OH)_2[/tex] solution contains the calculated number of moles of HCl. (I'm assuming you were given the molarity of [tex]Mg(OH)_2[/tex])

    M HCl x 1L soln / M [tex]Mg(OH)_2[/tex] = x L of soln

    convert L to mL, then you can multiply the answer by magnesium hydroxide's specific gravity (2.36) to get the amount in grams.


    number 2 is the same approach but the answer is left in mL.
     
  6. Oct 16, 2004 #5

    t!m

    User Avatar

    Is it wrong here to assume that the production of Magnesium Chloride and water is a "neutralization" of the HCl? It sounds like a relatively elementary [HS Chem or Intro to Chem] level question, so a balanced eqn. and mole ratios could find how many grams of magnesium hydroxide are needed to completely react with all .025 L of HCl ... thus ridding the acid. Too simplified?
     
  7. Oct 18, 2004 #6
    Question one:

    “How many grams of Mg(OH)2 (2 is a subscript) will be needed to neutralize
    25ml of stomach acid if stomach acid is 0.10 M HCl?”

    Display reaction equation (balanced) (Step 1)

    Mg(OH)2+2HCl-->MgCl2+2H2O

    1 to 2 ratio:

    25ml of 0.10 M (INITIAL CONCENTRATION BEFORE DILUTION) of HCl (Step 2)

    Concentration (M) = Number of moles / Volume

    Therefore number of moles = concentration x volume (Step 3)

    25ml = 0.025 dm^3

    Number of moles = 0.10 x 0.025 = 0.0025 moles of HCl

    Now back to Step 1, there is a ratio of 1:2, so we must divide answer by 2:

    Moles in ratio = 0.0025 / 2 = 0.00125 (Step 4)

    Now to get mass (amount in grams) (Step 5)

    Mass (w) = Number of moles x Mr (Molecular mass)

    Mr of Mg(OH)2 = 42.320g

    Therefore w = 0.00125 x 42.320 = 0.0529g

    So 0.0529g of Mg(OH)2 was needed to neutralise 0.0025 moles of Stomach acid (HCl).

    Question 2:

    “How many ml of a 0.10 NaOH solution are needed to neutralize 15ml of 0.20
    M H3PO4 (subscripts) solution???”

    H3PO4 + 3NaOH --> Na3PO4 + 3H2O (Step 1)

    As we can deduce from the above equation there is a 1:3 ratio (Step 2)

    Concentration (M) = Number of moles / Volume

    Therefore number of moles = concentration x volume (Step 3)

    15ml = 0.015 dm^3

    0.20 x 0.015 = 0.003 moles of H3PO4

    Since the ratio is 1:3 we must multiply the number of moles of H3PO4 by 3. (Step 4)
    Mole(3NaOH) = 0.003 x 3 = 0.009.

    Concentration (M) = Number of moles / Volume

    Therefore Volume = Concentration x Number of moles: (Step 5)

    Volume of (3NaOH) = 0.10 x 0.009 = 0.0009 dm^3

    Volume in ml = dm^3 x 1000

    Therefore volume of 3NaOH (in ml) = 0.0009 x 1000 = 0.9 ml of 3NaOH needed.
     
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