What mass of melamine, c3n3(nh2)3, will be obtained from 113.0 kg of urea, co(nh2)2, if the yield of the overall reaction is 73.0 % ?

The balanced equation for the reaction is:

6 HNCO(l) → C3N3(NH2)3(l) + 3 CO2(g)

Convert amount of urea from kg to moles

Molar Mass of urea = 60.06 g/mol so, 113 kg urea contains

113 kg / 60.06 = 1.88 mol urea

From balanced equation 6 moles of urea yields only 1 melamine, so divide the moles of urea by 6. 

1.88 / 6 = 0.313 kmol melamine

Now multiply 0.313 with molar mass of melamine that is 126 g/mol

126 x 0.313 = 39.438 kg

Yield of overall reaction is 73% so multiply 39.438 with 0.73

39.438 x 0.73 = 28.799 kg is the answer

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