Answered

zn (s) 2 hcl (aq) -----> zncl2 (aq) h2(g) if 520 ml of h2 is collected over water at 28oc and the atmospheric pressure is 1.0 atm, how many g of zn was used at the start of the reaction? (vapor pressure of water at 28oc is 28.3 mmhg)

Answer :

The grams of zinc used at the beginning of the reaction between Zn and HCl were 0.512 g

This mass of zinc reacted with hydrochloric acid under normal conditions and 520 ml of H2 were collected.

Using the Ideal Gas Equation

To calculate the moles of H2, the ideal gas equation PV = nrt is used

Where

P = Pressure

V = Volume

n = moles

T = temperature

R = Gas constant

And it clears up

n = PV / Rt

Calculation of moles of H2

Data

P = 28.3 mmHg = 0.372 atm

V = 520ml = 0.52L

n = ?

T = 28°C = 301.15K

R = 0.082 L atm/mol °K

n = PV / Rt

n = 0.372. 0.52L / 0.082 . 301.15

n = 0.19344 /24.6943

n = 0.0078 moles of H2

Moles of Zn used

According to the balanced equation of the reaction, 1 mol of Zn produces 1 mol of H2, therefore 0.0078 mol of Zn were used.

To know the grams of Zn, its molecular weight (65.38 g/mol) is used.

65.38g Zn/1 mol x 0.0078 mol = 0.512g

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