Question

Zinc (Zn) metal reacts with hydrochloric acid (HCl) according to the following reaction: Zn(s) + 2 HCl(aq) ZnCl2(aq) + H2(g)What volume of H2 gas would be collected over water if 6.93 g Zn metal were allowed to react with 25.0 mL of 1.0 M HCl at 25.0 oC? The vapour pressure of water at 25.0 oC is 23.8 torr, and the pressure of the laboratory is 1.00 atm

Answer 1

Zinc (Zn) metal reacts with hydrochloric acid (HCl) according to the following reaction:

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

Moles of HCl = 25 mL x 1 mol/L = 25*10^-3 moles of HCl

Mass of Zn = 6.93 g

Moles of Zn = mass / molecular weight = 6.93/65.38 =0.106 moles

1 mol of Zn reacts with 2 mol of HCl produces 1 mol of H2

0.106 moles of Zn reacts with (2*0.106) mol of HCl

But we have only 0.025 moles of HCl, therefore HCl is limiting reagent

1 mol of HCl reacts with 0.5 mol of Zn produces 0.5 mol of H2

0.025 mol of HCl reacts with 0.0125 moles of Zn produces 0.0125 mol of H2

when you collect a gas over water the gas mixture will contain water vapour. You need to determine the actual partial pressure of the collected gas in the mixture of gas and water vapour..

Partial pressure H2 = total pressure - vapour pressure H2 at 19 deg C
= 760.0 mmHg - 23.8 mmHg
= 736.2 mmHg

Then use ideal gas equaion to determine the volume of H2 gas

PV = nRT

P = pressure H2 = 736.2 mmHg
V = volume
n = moles = 0.0125
R = gas constant = 62.36367 mmHg L mol^-1 K^-1
T = temp in Kelvin = 298 K

n = PV / RT
0.0125 = 736.2 mmHg x V / 62.36367 x 298 K
V = 0.316 L

V = 316 mL

316 mL volume of H2 gas would be collected over water.