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)

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)