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A 0.333 g-sample of antacid was dissolved in 40.00 mL of 0.135 MHCI solution, then back-titrated to the end-point with 9.28 mL of a0.0203 M NaOH solution.a) Calculate number of moles of acid in the original 40.00 mL of HCI?b) How many moles of base were used in the back titation of excessHCI?c) How many moles of excess HCII?a How many moles of HCl reacted with the antacid sample?

Sagot :

A) To find the number of moles of a compound, we can use the Molarity formula, which tells us that molarity is equal to the number of moles divided by the volume in liters.

M = n/V

Now in our question we have a few informations, the molarity and the volume

0.135M = n/0.04L

Therefore the number of moles will be:

n = 0.135M * 0.04L

n = 0.0054 moles or 5.4 * 10^-3 moles

B) Using the same molarity equation as it was used in letter A, we can find the number of moles:

M = n/V

0.0203M = n/0.00928

n = 0.000188 moles or 1.88 * 10^-4

C) Since the ratio of this reaction HCl + NaOH is 1:1, which means I need 1 mol of HCl to react with 1 mol of NaOH, we can easily find the excess by subtracting both numbers of moles:

Moles of HCl - Moles of HCl that reacted

0.0054 - 0.00018 = 0.0052 of excess of HCl or 5.2 * 10^-3

HCl + NaOH -> NaCl + H2O

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