Question

Hydrogen gas can be prepared by reaction of zinc metal with aqueous HCl: Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g) How many liters of H2 would be formed at 636 mm Hg and 17 ∘C if 27.0 g of zinc was allowed to react? How many grams of zinc would you start with if you wanted to prepare 6.85 L of H2 at 284 mm Hg and 28.5 ∘C?

Answer 1

Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g)

from this balanced equation it is clear that

one mole of Zn will give one mole of H2 gas right?

now calculate the no ofmoles of Zn = weight of Zn / molar mass of Zn

= 27.0 g / 65.4 g /mol

= 0.413 mol

so 0.413 moles of Zn will give 0.413 moles of H2 gas

now use the formula

PV = nRT

P = pressure = 636 mm Hg convert in to atm = 0.837 atm

V = volume = need to calculte

T = temperature = 273 + 17 = 290K

n = no of moles = 0.413mol

R = 0.0821 Latm/mol-K

put all these values in the above equation

0.837 x V = 0.413 x 0.0821 x 290

V = 11.75 L

Part-2

calculate th eno o fmoles

P = 284 mm Hg = 0.374 atm

T = 273 + 28.5 = 301.5 K

V = 6.85L

n = we have to calculate

R = 0.0821 Latm/mol-K

0.374 x 6.85 = n x 0.0821 x 301.5

n = 2.57 / 24.75

n = 0.104 mole

from the balanced equation

0.104 moles of H2 will give 0.104 moles of Zn

now use the formula

moles = mass / molar mass

mass = 0.104 mol x 65.4 g/mol

mass = 6.8 gr