How many grams of precipitate will be formed when 20.5 ml of 0.800 m
co(no3)2 reacts with 27.0 ml of 0.800 m naoh in the following
chemical reaction?
co(no3)2 (aq) + 2 naoh (aq) → co(oh)2 (s) + 2 nano3 (aq)

The grams of precipitate formed when 20.5 mL of 0.800 M

CO(NO3)2 reacts with 27.0 mL of 0.800 M NaOH according to the reaction would be 1.98 g

From the balanced equation of the reaction:

The mole ratio of to is 1:2.

Mole of = molarity x volume

                                = 0.8 x 20.5/1000

                                 = 0.0164 moles

Mole of NaOH = 0.8 x 27/1000

                       = 0.0216 moles

Theoretically, the mole of NaOH should be 2x the mole of CO(NO3)2.

                                  0.0164 x 2 = 0.0328

Thus, NaOH is the limiting reagent.

Mole ratio of NaOH and CO(NO3)2 = 2:1

Therefore, mole of CO(NO3)2 = 1/2 x mole of NaOH

                                                  = 1/2 x 0.0216

                                                        = 0.0108 mole

Mass = mole x molar mass

Molar mass of CO(NO3)2 = 182.943 g/mol

Mass of CO(NO3)2 = 0.0108 x 182.943

                                = 1.98 g

More on calculating mass can be found here: link

There will be formed 1.84 grams of precipitate (NaNO3)

Explanation:

Step 1: The balanced equation

CO(NO3)2 (aq) + 2 NaOH (aq) → CO(OH)2 (s) + 2 NaNO3 (aq)

Step 2: Data given

Volume of 0.800 M  CO(NO3)2 = 20.5 mL = 0.0205 L

Volume of 0.800 M NaOH = 27.0 mL = 0.027 L

Molar mass of NaNO3 = 84.99 g/mol

Step 3: Calculate moles of CO(NO3)2

Moles CO(NO3)2  = Molarity * volume

Moles CO(NO3)2  = 0.800 M * 0.0205

Moles CO(NO3)2 = 0.0164 moles

Step 4: Calculate moles NaOH

moles of NaOH = 0.800 M * 0.027 L

moles NaOH = 0.0216 moles

Step 5: Calculate limiting reactant

For 1 mole CO(NO3)2 consumed, we need 2 moles of NaOH to produce 1 mole of CO(OH)2 and 2 moles of NaNO3

NaOH is the limiting reactant. It will completely be consumed.

CO(NO3)2 is in excess. There willbe 0.0216 / 2 = 0.0108 moles of CO(NO3)2 consumed. There will remain 0.0164 - 0.0108 = 0.0056 moles of CO(OH)2

Step  6: Calculate moles of NaNO3

For 2 moles of NaOH consumed, we have 2 moles of NaNO3

For 0.0216 moles of NaOH, we have 0.0216 moles of NaNO3

Step 7: Calculate mass of NaNO3

mass of NaNO3 = moles of NaNO3 * Molar mass of NaNO3

mass of NaNO3 = 0.0216 moles * 84.99 g/mol = 1.84 grams

There will be formed 1.84 grams of precipitate (NaNO3)

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