Question

How many moles of HCl are required to react with 1 g of zinc?

Answer 1

Answer:

Explanation:

To solve problems like this, the very first thing to do is to write out the balanced equation for the process. The reaction is

Zn(s) + HCl(aq) → ZnCl2(aq) + H2(g)

and the balanced equation is

Zn(s) + 2 HCl(aq) → ZnCl2(aq) + H2(g)

So, by examining the stoichiometry, you can see that for every mole of Zn, two moles of HCl will be required. In turn, we need to know the number of moles of Zn is represented by 10.4 g. The molecular weight of Zn is 65.38 g/mol so

moles of Zn = 10.4 g/65.38 g/mol = 0.159 moles

and, twice as many moles of HCl is required so

# of moles of HCl required = 2 * 0.159 = 0.318 moles