A 9.03 g sample of an aqueous solution of nitric acid contains an unknown amount of the acid. If 12.0 mL of 0.655 M barium hydroxide are required to neutralize the nitric acid, what is the percent by mass of nitric acid in the mixture?

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Olajidey Olajidey · Answer 1 · 2020-03-16T03:32:06+01:00

Answer:

The percent by mass of nitric acid in the mixture is 5.48 %

Explanation:

Givn that

Mass of HNO3 = 9.03 grams

Volume of KOH = 12.0 mL = 0. 012 L

Molarity of KOH = 0.655 M

The balanced equation

Ba(OH)2 + HNO3 → Ba(NO3)2 + H2O

Calculate the moles of KOH

Moles of Ba(OH)2 = molarity KOH * volume

Moles Ba(OH)2 = 0.655 M * 0.012 L

Moles Ba(OH)2 = 0.00786 moles

Calculate moles of HNO3

For 1 mol of Ba(OH)2 we need 1 mol of HNO3

For 0.00786 moles of Ba(OH)2 we need 0.00786 moles of HNO3

Calculate mass of HNO3

Mass HNO3 = moles HNO3 * molar mass HNO3

Mass HNO3 = 0.00786 moles * 63.01 g/mol

Mass HNO3 = 0.495 grams

Calculate mass % HNO3 in sample

mass % = (0.495 grams / 9.03 grams)*100%

mass % = 5.48 %

The percent by mass of nitric acid in the mixture is 5.48 %