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24 June, 08:26

A 8.72 g sample of an aqueous solution of hydrobromic acid contains an unknown amount of the acid. If 19.5 mL of 0.374 M barium hydroxide are required to neutralize the hydrobromic acid, what is the percent by mass of hydrobromic acid in the mixture?

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  1. 24 June, 08:52
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    3.3 %

    Explanation:

    According to the question, the following reaction takes place -

    Ba (OH) ₂ + 2 HBr → BaBr₂ + 2 H₂O

    Molarity of a substance, is the number of moles present in a liter of solution.

    M = n / V

    M = molarity

    V = volume of solution in liter,

    n = moles of solute,

    According to the question,

    V = volume of Ba (OH) ₂ = 19.5 mL = 0.0195 L (since, 1 ml = 10 ⁻³ L)

    M = Molarity of Ba (OH) ₂ = 0.374 M

    The moles of Ba (OH) ₂ can be calculated by using the above equation,

    M = n / V

    n = M * V = 0.374 M * 0.0195 L = 0.0072 mol

    From the above balanced reaction,

    2 mol of HBr reacts with 1 mol Ba (OH) ₂

    1 mol of HBr reacts with 1 / 2 mol Ba (OH) ₂

    From the above data,

    1 mol HBr reacts with = 1 / 2 * 0.0072 mol = 0.0036 mol

    Hence, number of moles of HBr = 0.0036 mol

    Now,

    Moles is denoted by given mass divided by the molecular mass,

    Hence,

    n = w / m

    n = moles,

    w = given mass,

    m = molecular mass.

    As calculated above,

    n = 0.0036 mol

    As we know, the m = molecular mass of HBr = 81 g/mol

    n = w / m

    w = n * m = 0.0036 mol * 81 g/mol = 0.2916 g

    Now,

    mass % = mass of HBr / mass of solution * 100

    mass % = 0.2916 g / 8.72 g * 100 = 3.3 %
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