What volume (in mL) of 3.2 M hydrobromic acid is required to entirely dissolve 19.4 g of lithium metal?

69.7 mL ≅ 70 mL

Explanation:

The chemical reaction between hydrobromic acid (HBr) and lithium metal (Li) produces lithium bromide (LiBr) and hydrogen (H₂) according to the following equation:

2HBr(aq) + 2Li(s) → 2LiBr(aq) + H₂(g)

Thus, 2 moles of HBr reacts with 2 moles of Li. So, the stoichiometric ratio is 1:1 (1 mol HBr reacts with 1 mol Li).

We convert the mass of Li to moles by using the molar mass (MM) of LiBr:

MM(LiBr) = 6.9 g/mol + 79.9 g/mol = 86.8 g/mol

moles of Li(s) = mass/MM(LiBr) = 19.4 g/86.8 g/mol = 0.223 moles Li(s)

Now, we can divide the moles of Li into the molarity of the HBr solution to obtain the volume in liters wee need to add:

V = moles Li(s)/M = (0.223 mol)/(3.2 mol/L) = 0.0697 L

0.0697 L x 1000 mL/1 L = 69.7 mL ≅ 70 mL

Therefore, 70 mL of 3.2 M HBr are needed to entirely dissolve the metal.

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