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2 May, 08:19

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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  1. 2 May, 10:11
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    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 %
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