Question

How many grams of solid barium hydroxide are needed to react with 10.5 mL of a 0.473 M nitric acid solution? Assume that the volume remains constant when the barium hydroxide is added.

Ba(OH)2 + 2HNO3Ba(NO3)2 + 2 H2O

Answer 1

The reaction is

Ba(OH)₂ + 2HNO₃$\rightarrow$ Ba(NO₃)₂ + 2H₂O

which shows that when 1 mole of barium hydroxide is reacts with 2 moles of nitric acid it gives product Ba(NO₃)₂

let us find moles of nitric acid with the given data

We know that

Molarity = moles / Liter of solution

As volume remains same , therefore volume of solution = 10.5 ml = 0.0105 L

therefore

o.473 M = moles of Nitric acid / 0.0105 L

then

Moles of nitric acid = 0.473 M x 0.0105 L = 4.966 x 10^-3 moles

therefore

moles of Ba(OH)₂ = 4.996  x 10^-3 /2 = 2.48  x 10^-3 moles

So

2.48 x 10^-3 moles of Ba(OH)₂ reacts with 4.96 x 10^-3 mole of nitric acid to form product

Now

we know that

Mole = mass / molar mass

then

Mass of Ba(OH)₂ = moles of Ba(OH)₂ x molar mass of Ba(OH)₂

= 2.48 x 10^-3  moles x 171.34 g/mol

= 0.43 g

therefore 0.43 g of Ba(OH)₂ is needed