Question

If 5.70mL of 0.1000 M NaOH solution is needed to just neutralize excess acid after 20.00 mL of 0.1000 M HCl was added to 1.00 g of an antacid, how many moles of acid can the antacid counteract per gram?

Answer 1

NaOH + HCl => NaCl + H2O

Moles of excess HCl = moles of NaOH

= volume x concentration of NaOH

= 5.70/1000 x 0.1000 = 0.00057 mol

Initial moles of HCl = volume x concentration of HCl

= 20.00/1000 x 0.1000 = 0.002 mol

Moles of acid that the antacid counteract per gram

= moles of reacted HCl

= initial moles of HCl - moles of excess HCl

= 0.002 - 0.00057

= 0.00143 mol = 1.43 x 10^-3 mol

Answer 2

5.25 mL *0.1 mol OH-/L * (1L/1000 mL ) = 0.000525 moles OH-

0.000525 moles of OH- react with the excess H+ from HCl. We canthen calculate the amount of total H+ added (the excess reactedwith OH-, but the other amount reacted with antacid)

0.1 mol H+/L *(20 mL) *(1L/1000mL) = 0.002 moles of H+total

Thus the antacid reacted with 0.002 - 0.000525 = 0.001475 molesH+

Moles of acid counteracted per gram of antacid

0.001475 moles/1 g = 0.001475 moles/g

Are you sure you want g/mol? The answer would make more sense inmoles/gram