What is the minimum volume of 3.0 M HBr required to completely dissolve 15.0 g of potassium metal?

Solution 92P :

Step 1:

Given :

Molarity of HBr (Hydrobromic acid) = 3.0 M

Amount of K(Potassium) metal = 15.0 g

Volume of HBr to completely dissolve Mg = ?

Let’s write the chemical equation for the given scenario :

2 K(s) + 2HBr → 2 KBr + H2(g)

From the above reaction, we see the ratio of reactants is 2 : 2.

Now, let’s find the number of moles of K and HBr :

Number of moles(n) =

Molar mass of K = 39.0983 g/mol.

Therefore, n =

= 0.3836 moles.

We are not given the mass of HBr, but from the reaction, we know that the ratio of K and HBr is 2 : 2. Hence, the moles of HBr will be 0.3836 moles.