Zinc reacts with hydrochloric acid according to the reaction equation Zn ( s ) + 2 HCl ( aq ) ⟶ ZnCl2 ( aq ) + H 2 ( g ) How many milliliters of 5.00 M HCl ( aq ) are required to react with 8.05 g Zn ( s )


Answer:

The volume for the reaction is 49.2 mL

Explanation:

We define the reaction where 1 mol of solid Zn reacts with 2 moles of hydrochloric to produce zinc chloride and hydrogen

The equation is:

Zn (s) + 2 HCl (aq) ⟶ ZnCl₂ (aq) + H₂(g)

First of all, we convert the mass of Zn to moles → 8.05 g / 65.41 g/mol = 0.123 moles

Ratio is 1:2. Then 0.123 moles will need the double of moles of HCl to react. (0.123 . 2 ) = 0.246 moles

We do not know the volume, but we have the molarity of the acid.

Molarity = moles of solute / volume (L) of solution

Therefore, volume (L) of solution = moles of solute / Molarity

Volume (L) of solution = 0.246 moles / 5 mol/L = 0.0492 L

0.0492L = 49.2 mL


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