What is the volume (in mL) of a 0.265 M HCl solution required to completely react with 65.0 grams of Fe?

Question

Fe (s) + 2 HCl (aq) 🡪 FeCl2 (aq) + H2 (g)  

Answer 1

Fe(s) + 2HCl(aq) ==> FeCl2(aq) + H2(g) ... balanced equation

How many moles of Fe do we have?

** 65.0 g Fe x 1 mol Fe/55.8 g = 1.165 moles Fe

How many moles of HCl do we need to react with this amount of Fe?

** 1.165 moles Fe x 2 mol HCl/mol Fe = 2.33 moles HCl (this is based on the 2:1 mol ratio in balanced eq)

How many mls of 0.265 M HCl do we need to obtain 2.33 moles HCl?

** (x L)(0.265 mol/L) = 2.33 moles and x = 8.79 L = 8790 mls