how many milliliters of a 0.46 M HCl solution are needed to react completely with a 6.4 g of zinc to form zinc (ii) chloride answer in mL

Help please :( I was absent and don't know where to start

1 answer

start by reading your text book. I'm sure there's a section regarding moles.

Zn has a mol wt of 65.39, so 6.4g = 0.1 moles of Zn

Since the reaction is

Zn + 2HCl = ZnCl2 + H2
each mole of Zn reacts with 2 moles HCl

So, you need .2 moles HCl

.46M HCl has .46 moles/liter

So, you need .2/.46 = .434 liters, or 434ml