# Chemistry CHALLENGED

1. Oct 13, 2004

### TJSR

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. Oct 13, 2004

### maverick280857

3. Oct 14, 2004

### wolfson_1123

All i can say if please use:

Ma x Mb / Va x Vb transpose for what you need.

4. Oct 15, 2004

### jreason

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 $$Mg(OH)_2$$ solution contains the calculated number of moles of HCl. (I'm assuming you were given the molarity of $$Mg(OH)_2$$)

M HCl x 1L soln / M $$Mg(OH)_2$$ = 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.

5. Oct 16, 2004

### t!m

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?

6. Oct 18, 2004

### wolfson_1123

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.