What is the total number of grams of h20 produced when 116 grams of the product, naci, is formed?
a) 36 9g
b) 54 9
c) 18 g
d) 9.0 g

A) 36 g

Explanation:

From the source, some values have been correct.

Given that;-

Mass of NaCl formed = 116 grams

Molar mass of NaCl = 58 g/mol

Moles of NaCl =

According to the reaction shown below:-

1 mole of NaCl and 1 mole of are produced upon reaction.

So,

2 moles of NaCl and 2 moles of are produced upon reaction.

Moles of water = 2 moles

Molar mass of water = 18 g/mol

Mass =

36 g of is produced.

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