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?

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 %

b is the correct answer

mark me brainliest and hope this

може ли да зареждате повече информация, моля