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)

b is the correct