In 2Fe+3Cl2=2FeCl3, how much iron (Fe) is needed to produce 4 moles of FeCl3? In 2Fe+3Cl2=2FeCl3, how much FeCl3 is produced when 5 moles of Fe react in excess chlorine?​

4 mol Fe

5 mol FeCl₃

Explanation:

Step 1: Write the balanced equation

2 Fe + 3 Cl₂ ⇒ 2 FeCl₃

Step 2: Calculate how much iron (Fe) is needed to produce 4 moles of FeCl₃

According to the balanced equation, the molar ratio of Fe to FeCl₃ is 2:2.

4 mol FeCl₃ × 2 mol Fe/2 mol FeCl₃ = 4 mol Fe

Step 3: Calculate how much FeCl₃ is produced when 5 moles of Fe react in excess chlorine

According to the balanced equation, the molar ratio of Fe to FeCl₃ is 2:2.

5 mol Fe × 2 mol FeCl₃/2 mol Fe = 5 mol FeCl₃

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