Achemist needs to find out how large of a container would be required to contain the products of an ammonia decomposition. if 8.91 mol h2 are produced during the reaction, how many moles of n2 are produced?

Answer : The moles of produced are, 2.97 moles

Solution : Given,

Moles of = 8.91 moles

The balanced decomposition reaction of ammonia will be,

From the balanced reaction, we conclude that

When 2 moles of decomposes to produce 3 moles of and 1 mole of

If the moles of is 3 moles then the moles of produces = 1 mole

If the moles of is 8.91 moles then the moles of produces =

Therefore, the moles of produced are, 2.97 moles

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

lol idk man i guess u gotta ask bill nye the science guy