Bl. how many mole of alo, (aluminum oxide) would be produced from reacting 6 mole of feo?

2 mol of Al₂O₃ are produced from 6 mol of FeO

Explanation:

The reaction to produce aluminum oxide from FeO is this:

3FeO  +  2Al  → 3Fe  + Al₂O₃

Ratio is 3:1

3 mol of FeO are necessary to produce 1 mol of aluminum oxide

6 mol of FeO will produce (6 .1)/3 = 2 moles

There are 2.0 moles of Al2O3 produced

Explanation:

Step 1: Data given

Number of moles of FeO = 6.0 moles

Step 2: The balanced equation

2Al + 3FeO → Al2O3 + 3Fe

Step 3: Calculate moles of Al2O3

When 2 moles of Al are consumed, we need 3 moles of FeO to produce 1 mol of Al2O3 and 3 moles of Fe

When 6 moles of FeO are consumed, we'll have 6.0 / 3 = 2.0 moles of Al2O3

There are 2.0 moles of Al2O3 produced

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

the temperature of the gas, in k, not celsius.