Question

If a 0.4681 g Mg strip reacts with 0.650 M HCl in a 139.3 mL flask at 25oC, what is the minimum volume (mL) of HCl needed to completely react with the magnesium?

Answer 1

The reaction between the magnesium, Mg, and the hydrochloric acid, HCl is given in the equation below,

    Mg + 2HCl --> H2 + MgCl2

The number of moles of HCl that is needed for the reaction is calculated below.
    n = (0.4681 g Mg)(1 mol Mg/24.305 g Mg)(2 mol HCl/1 mol Mg)
    n = 0.0385 mols HCl

From the given concentration, we calculate for the required volume. 
    V = 0.0385 mols HCl/(0.650 mols/L)
     V = 0.05926 L or 59.26 mL

Answer: 59.26 mL of HCl