What mass of Na2SO4 will be formed by addition of 17.0 g of NaHCO3 in aqueous solution to an aqueous solution containing 0.400 mol of H2SO4?

Mass of Na₂SO₄= 14.346 g

Given

17 g NaHCO₃

0.4 mol H₂SO₄

Required

mass of Na₂SO₄

Solution

Reaction

H₂SO₄ (aq) + 2 NaHCO₃ (aq) ⇒ Na₂SO₄(aq) + 2 CO₂(g) + 2 H₂O (l)

mol NaHCO₃ :

= mass : MW

= 17 g : 84 g/mol

= 0.202

Find limiting reactant = mol : coefficient

H₂SO₄ = 0.4 :  1 = 0.4

NaHCO₃ = 0.202 : 2 = 0.101

NaHCO₃ as a limiting reactant (smaller ratio)

Mol Na₂SO₄ based on mol NaHCO₃

From the equation, mol ratio NaHCO₃ : Na₂SO₄ = 2 : 1, so mol Na₂SO₄ :

= 1/2 x mol NaHCO₃

= 1/2 x 0.202

= 0.101

Mass of Na₂SO₄ :

= mol x MW

= 0.101 x 142,04 g/mol

= 14.346 g

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years