When copper (ii) nitrate is mixed with sodium hydroxide, a blue precipitate, (copper (ii) hydroxide) forms. the other product is sodium nitrate. when 15.25 grams of copper (ii) nitrate is reacted with 12.75 grams of sodium hydroxide, what is the percent yield of the precipitate if 5.23 grams are recovered?

percentage yield = 67%

Explanation:

Given data:

Mass of Cu(NO₃)₂  = 15.25 g

Mass of NaOH   = 12.75 g

Percentage yield = ?

Solution:

Cu(NO₃)₂ + 2NaOH   →  Cu(OH)₂ + 2NaNO₃

Moles of Cu(NO₃)₂:

Number of moles = mass/ molar mass

Number of moles = 15.25 g /187.56 g/mol

Number of moles = 0.08 mol

Moles of NaOH :

Number of moles = mass/ molar mass

Number of moles = 12.75 g / 40 g/mol

Number of moles = 0.32 mol

Now we will compare the moles of Cu(OH)₂ with NaOH and Cu(NO₃)₂.

                             NaOH             :      Cu(OH)₂

                                2                   :          1

                                0.32              :           1/2×0.32 = 0.16 mol

                             Cu(NO₃)₂         :           Cu(OH)₂

                                   1                  :               1

                              0.08                :              0.08

The number of moles produced by  Cu(NO₃)₂  are less so it will limiting reactant.

Mass of Cu(OH)₂:

Mass = number of moles × molar mass

Mass = 0.08 mol × 97.6 g/mol

Mass = 7.808 g

Theoretical yield = 7.808 g

Percent yield:

percentage yield = Actual yield/ theoretical yield ×  100

percentage yield = 5.23 g/  7.808 g ×  100

percentage yield = 0.67 ×  100

percentage yield = 67%

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

c. heat could easily travel through the molecules in the air and escape/enter the flask. removing the air slows down the heat transfer.