A8.72 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?

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 %

i'm not sure

explanation:

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