Answer:
There will be formed 1.84 grams of precipitate (NaNO3)
Explanation:
<u>Step 1</u>: The balanced equation
CO(NO3)2 (aq) + 2 NaOH (aq) → CO(OH)2 (s) + 2 NaNO3 (aq)
<u>Step 2:</u> 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
<u>Step 3:</u> 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)
