Question

How many grams of Mg(OH)2 will be needed to neutralize 25mL of stomach acid if stomach acid is .10M HCL?

How many mL of a .10 M NaOH solution are needed to neutralize 15mL of 0.20 M H3PO4 solution?

Answer 1

How many mL of a .10 M NaOH solution are needed to neutralize 15mL of 0.20 M H3PO4 solution?

The formula is M1 V1 = M2 V2

M1 = 0.10 M, V1 = ? M2 = 0.20 M, V2 = 15 mL

0.10 X V1 = 0.20 X 15

V1 = 0.20 X 15 / 0.10

= 30 mL

How many grams of Mg(OH)2 will be needed to neutralize 25mL of stomach acid if stomach acid is .10M HCL?

We need to find out how much HCl is in the stomach acid. Using 25 mL stomach acid x [0.1 moles HCl/1000 mL stomach acid) gives us 0.0025 moles of HCl.

Balancing the equation for the neutralization reaction, we get

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

or 2 moles of HCl react with 1 mole of Mg (OH)2

so 0.0025 moles of HCl will react with half as many moles of Mg(OH)2, written:

0.0025 moles HCl x (1 mole Mg(OH)2/2 moles HCl) = 0.00125 moles Mg(OH)2.

To calculate the number of grams of Mg(OH)2, use the molar mass of Mg(OH)2:

0.00125 moles Mg(OH)2 x 58.05 g Mg(OH)2/mole Mg(OH)2 =
0.0726 g of Mg(OH)2

It looks like the problem included only 2 significant figures, so the answer would be rounded to 0.073 g of Mg(OH)2