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15 June, 14:13

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?

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  1. 15 June, 15:13
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    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
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