4p + 5O2 -> P4O10 ; the percent yield of PO4O10 when 6.20 g of phosphorus burns into excess oxygen is 67.0%. What is the yield of P4O10?

   36.9 g

Explanation:

P4 + 5O2 = P4O10  Balanced equation

moles P4 present = 23.9 g x 1 mole/123.88 g = 0.193 moles

moles O2 present = 20.8 g x 1 mol/32 g = 0.65 moles O2

From balanced equation, mole ratio O2 : P4 is 5:1.  Is 0.65 moles O2 5x 0.193 moles?  NO.  You don't have enough O2.

O2 is limiting in this reaction.

theoretical moles of P4O10 = 0.65 moles O2 x 1 mole P4O10/5 moles O2 = 0.13 moles P4O10

mass of P4O10 produced = 0.13 moles x 283.9 g = 36.9 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