Zn reacts with hydrochloric acid. Zn(s)+ 2 HCl(aq)$\to$ ZnCl₂(aq)+ H₂(g) What volume of 3.05 M HCl(aq)will react with 25.0 g Zn(s)?

Question

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First, calculate moles of Zn: 25.0 g Zn / 65.38 g/mol = 0.382 mol Zn. The reaction requires 2 moles of HCl per mole of Zn, so 0.382 mol Zn needs 0.764 mol HCl. Using the concentration of HCl (3.05 M), volume = moles/concentration = 0.764 mol / 3.05 M = 0.251 L.

Zn Reacts with Hydrochloric Acid →\to→ ZnCl2(aq)+ H2(g) What Volume of 3

Zn Reacts with Hydrochloric Acid $\to$ ZnCl₂(aq)+ H₂(g)
What Volume of 3