Question

11. Zinc reacts with aqueous sulfuric acid to form hydrogen gas: Zn(s) + H2SO4(aq) → Z1804(aq) + H2(g) In an experiment, 201 ml of wet H2 is collected over water at 27°C and a barometric pressu of 732 torr. How many grams of Zn have been consumed? The vapor pressure of water at 27° is 26.74 torr. a. 4.18 x 106 g b. 0.496 g c. 496 g d. 377 g e. 3.77 x 105 g

Answer 1

ANSWER : b. 0.496 g

Given :

volume of wet hydrogen gas = 201 ml = 0.201 L

Temperature of a gas = 27 ⁰ C = 300.15 K

Pressure of wet hydrogen gas = 732 torr

vapor pressure of water at 300 K = 26.74 torr

We know that, when we collect gas over water it is not pure gas. It is a mixture of gas & water vapor. Hence, we can write P total = P gas + P water vapor.

$\therefore$ P gas = P total - P water vapor = 732 torr - 26.74 torr = 705.26 torr

i e Pressure of dry H 2 gas = 705.26 torr

We have relation, 1 atm = 760 mm Hg.

Hence, Pressure of Dry H ₂ gas =705.26 torr $\times$ (1 atm / 760 torr ) =0.9279 atm

.We have relation, P V = n R T

Where P is pressure of a gas, V is a volume of a gas , n is no. of moles of gas and T is temperature of gas.

Therefore, number of moles of H ₂ gas = P V / R T

No. of moles of H ₂ gas = 0.9279 atm $\times$ 0.201 L / 0.082057 L atm / mol K $\times$ 300 .15 K

No. of moles of H ₂ gas =0.00757 mol

Consider given reaction, Zn (s) + H₂SO₄ (aq)  $\rightarrow$ ZnSO₄ (aq) + H ₂ (g)

From reaction, 1 mol Zn $\equiv$ 1 mol H₂SO₄$\equiv$ 1 mol  ZnSO₄$\equiv$ 1 mol H ₂

i e No.of moles of Zn metal reacted = No. of moles of hydrogen gas produced.

$\therefore$ No.of moles of Zn metal reacted = 0.00757 mol

We know that, No. of moles = Mass / molar mass

$\therefore$ Mass = No. of moles $\times$ molar mass

$\therefore$ Mass of Zn metal consumed = 0.00757 mol $\times$ 65.37 g/mol = 0.495 g

Our answer is close to value 0.496 g . Hence , answer is 0.496 g.

ANSWER : b. 0.496 g