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?
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
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