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 742 mm Hg and 15 °C if 25.5 g of zinc was allowed to react?

Answer 1

Answer:

The volume of hydrogen gas generated by given reaction is 9.48 L

Explanation:

Firstly, we calculate the mole of Zinc added, n = 25.5 / 65 = 0.392 mol

With the same equivalent, the mole of generated hydrogen gas will be the same, at 0.392 mol.

Secondly, we get the equation

P . V = n . R . T

with P (pressure, atm), V (volume, litre), n (mole, mol), R (ideal gas constant, 0.082), T (temperature, Kelvin)

So we have to convert the given information to correct unit

P = 742 mmHg = 0.976 atm

T = 15 °C = 288 K

Hence we can calculate the generated volume by

V = n . R . T / P = 0.392 x 0.082 x 288 / 0.976 = 9.48 L