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Zinc metal reacts with HCl as follows: Zn + 2 HCl ⇒ ZnCl2 + H2 What mass of zinc metal is needed to produce 600.0 mL of hydrogen gas collected over water at 30.0°C and at 775 mmHg?

User IgorL
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1 Answer

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Answer:

it is required a mass of Zn equal to m= 1.608 gr of Zn

Step-by-step explanation:

Assuming that the hydrogen behaves as an ideal gas ( in reality it has positive deviations from the gas law, but we can neglect the small correction) , then

P*V=n*R*T

n = P*V/(R*T)

where

P= absolute pressure = 775 mmHg = 1.02 atm

T= absolute temperature = 30°C+273 = 303 K

V= volume = 600 mL = 0.6 L

R = ideal gas constant = 0.082 atm*L/(mol*K)

n = number of moles of H₂ gas

replacing values

n = P*V/(R*T) = 1.02 atm*0.6 L/(0.082 atm*L/(mol*K)*303 K) = 0.0246 moles of H₂

from the reaction

Zn + 2 HCl ⇒ ZnCl2 + H2

since 1 mole of H2 requires 1 mole of Zn to react , then

m = nz * M

where

nz = number of moles of zinc = number of moles of H₂ gas = 0.0246 moles

M = molecular weight of zinc = 65.38 gr/mol

m = mass of zinc required

therefore

m= 0.0246 moles * 65.38 gr/mol = 1.608 gr of Zn

m= 1.608 gr of Zn

User Colibrow
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