Question

Passage 14 (Questions 78–81)

The following experiments study the reactivities of several active metals (Experiment 1) and halogens (Experiment 2).

Experiment 1

Pea-sized samples of five active metals were placed in deionized water, and observations were recorded in Table 1.

Table 1

Observations

Metal used	Observation upon adding the metal to water	Observation of the resulting solution
Mg	No obvious reaction	Neutral
Ca	Sank, and slowly gave off bubbles of a gas	Basic
Li	Moved over the surface slowly fizzing	Basic
Na	Moved over the surface vigorously fizzing, caught fire	Basic
K	Moved over the surface vigorously fizzing, exploded loudly	Basic

Experiment 2

A saturated aqueous solution of Cl₂ was added to separate aqueous solutions of NaF, NaCl, NaBr, and NaI and mixed well. Observations were recorded. In addition, samples of salt solutions were added to separate electrolysis cells, and the minimum voltage required to produce an observable reaction at the anode was recorded. H₂(g) was produced at the cathode in each cell and, except for the NaF solution, the solutions became basic. Results are recorded in Table 2.

Table 2

Result

Solution	Chlorine water	Electrolysis cell voltage and product
NaF	No change	2.06 V; O2(g)
NaCl	No change	2.19 V; Cl2(g)
NaBr	Red-brown	1.90 V; Br2(aq)
NaI	Yellow-brown	1.37 V; I2(aq)

Experiment 1 was repeated with 0.40 g of calcium, and the gas that evolved was collected. The identity of the gas, and its approximate volume at 1.0 atm and 27°C were:
(Note: R = 0.0821 L•atm/mol•K)

H₂, 250 mL.

H₂, 500 mL.

O₂, 250 mL.

O₂, 500 mL.

Answer 1

Reaction of Calcium metal with water forms calcium hydroxide with the liberation of hydrogen gas the reaction be writen as

Ca + H₂O = Ca(OH)₂ + H₂

For 1 mole of calcium 1 mole of Hydrogen gas is liberated.

For 0.4 g of Calcium metal is used

Number of moles of calcium = weight of calcium used/ atomic mass of calcium

                = 0.4 / 40 = 0.01

0.01 moles of Calcium is used, which liberates the 0.01 moles of hydrogen gas (H₂)

Ideal gas equation can be used to find out the volume of hydrogen gas formed or liberate during the reaction.

PV = nRT

Given values:

pressure P = 1 atm

V = ? volume of hydrogen gas has to determined in Litre

n = 0.01 mole of hydrogen liberated

R = 0.0821 L.atm/mol.K

T = 27 ⁰C = 273 + 27 = 300 K converted to Kelvin SI unit.

On substituting all values

1 X V = 0.01 X 0.0821 X 300

V = 0.2463 L or 0.2463 litre

V = 0.2463 X 1000 = 246.3 mL (converted to mL by multiplying with 1000)

Approximately the volume V = 250 mL of H₂

First option is correct.