An iron bar weighed 565 g. After the bar had been standing in moist air for a month, exactly one-eighth of the iron turned to rust (Fe2O3). Calculate the final mass of the iron bar and rust.

Final mass of Iron bar will be 621.105 g.

Mass of rust = 126.730 g

Explanation:

Exactly one- eighth of the Iron turns to rust. That means, the mass of Iron  that under go the process of rusting() = 565 × = 70.625 g

So, the mass of pure Iron after rusting will be = 565 - 70.625 = 494.375 g

Now the reaction of rusting as follows,

   2Fe + →

Now, molecular weight of Iron = 55.84 g

So, moles of Iron undergoes the reaction is = = 1.26 mol

Molecular weight of = 159.68 g

Here 2 moles of Iron undergoes the reaction to produce 2 moles of Iron oxide.

Hence, 1.26 mol of Iron will produce 1.26 mole of Iron oxide.

Mass of = = 126.730 g

∴Mass of rust = 126.730 g

∴ Final mass of Iron bar = ( 494.375 + 126.730) = 621.105 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