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?

Question

Jensen Moore

Chemistry

7 September, 02:46

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?

+1

Answers (1)

Know the Answer?

Theresa Weber · 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