If a solution containing 53.99 g 53.99 g of mercury(II) nitrate is allowed to react completely with a solution containing 9.718 g 9.718 g of sodium sulfide, how many grams of solid precipitate will form?

28.97 g

Explanation:

The formula for the calculation of moles is shown below:

For mercury(II) nitrate

Mass = 53.99 g

Molar mass = 324.7 g/mol

The formula for the calculation of moles is shown below:

Thus,

For sodium sulfide

Mass = 9.718 g

Molar mass = 78.0452 g/mol

The formula for the calculation of moles is shown below:

Thus,

According to the given reaction:

1 mole of mercury(II) nitrate reacts with 1 mole of sodium sulfide

So,  

0.16627 mole of mercury(II) nitrate reacts with 0.16627 mole of sodium sulfide

Moles of sodium sulfide = 0.16627 moles

Available moles of sodium sulfide = 0.12451 moles

Limiting reagent is the one which is present in small amount. Thus, sodium sulfide is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of sodium sulfide produces 1 mole of mercury sulfide (precipitate)

So,  

0.12451 mole of sodium sulfide produces 0.12451 mole of mercury sulfide (precipitate)

Moles of mercury sulfide  = 0.12451 mole

Molar mass of mercury sulfide  = 232.66 g/mol

Mass of mercury sulfide  = Moles × Molar mass = 0.12451 × 232.66 g = 28.97 g

28.97 g  of solid precipitate will form.

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

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