Question

1.

A) How many grams of Mg(OH)₂ will be needed to neutralize 25mL of stomach acid if stomach acid is 0.10 M HCl?

B) How many mL of a 0.10 M NaOH solution are needed to neutralize 15 mL of 0.20 M H₃PO₄ solution?

Answer 1

A)

Balanced chemical equation is:

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

lets calculate the mol of HCl

volume , V = 25 mL

= 2.5*10^-2 L

use:

number of mol,

n = Molarity * Volume

= 0.1*2.5*10^-2

= 2.5*10^-3 mol

According to balanced equation

mol of Mg(OH)2 reacted = (1/2)* moles of HCl

= (1/2)*2.5*10^-3

= 1.25*10^-3 mol

This is number of moles of Mg(OH)2

Molar mass of Mg(OH)2,

MM = 1*MM(Mg) + 2*MM(O) + 2*MM(H)

= 1*24.31 + 2*16.0 + 2*1.008

= 58.326 g/mol

use:

mass of Mg(OH)2,

m = number of mol * molar mass

= 1.25*10^-3 mol * 58.33 g/mol

= 7.291*10^-2 g

Answer: 7.29*10^-2 g

B)

Balanced chemical equation is:

H3PO4 + 3 NaOH ---> Na3PO4 + 3 H2O

Here:

M(H3PO4)=0.2 M

M(NaOH)=0.1 M

V(H3PO4)=15.0 mL

According to balanced reaction:

3*number of mol of H3PO4 =1*number of mol of NaOH

3*M(H3PO4)*V(H3PO4) =1*M(NaOH)*V(NaOH)

3*0.2 M *15.0 mL = 1*0.1M *V(NaOH)

V(NaOH) = 90 mL

Answer: 90 mL