Question

Hydrogen gas can be prepared by reaction of zinc metal with aqueous HCl :
Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g)

1)How many liters of H2 would be formed at 592 mm Hg and 17 ∘C if 25.5 g of zinc was allowed to react?

2)How many grams of zinc would you start with if you wanted to prepare 4.60 L of H2 at 294 mm Hg and 34.5 ∘C ?

Answer 1

   Zn(s)+2HCl(aq) --- > ZnCl2(aq)+H2(g)

1 mol Zn = 1 mol H2

1) no of mol of Zn reacted = 25.5/65.4 = 0.4 mol

   no of mol of H2 liberated = 0.4 mol

   volume of H2 liberated = nRT/P

                           = 0.4*0.0821* 290.15 / (592/760)

                           = 12.23 L

2) no of mol of H2 leberated = PV/RT

                               = (294/760)*4.6/(0.0821*307.65)

                               = 0.07 mol

     no of mol of Zn must be reacted = 0.07 mol

     mass of Zn reacted = n*M

                         = 0.07*65.4

                         = 4.6 g