Final answer:
Approximately 1.525 grams of precipitate will be formed when 20.5 mL of 0.800 M Co(NO₃)₂ reacts with 21.0 mL of 0.800 M NaOH.
Step-by-step explanation:
To calculate the mass of precipitate formed, we need to find the limiting reactant in the reaction. The limiting reactant is the one that is completely consumed, determining the amount of product formed. To find the limiting reactant, we compare the moles of each reactant using the balanced equation. In this case, we have 20.5 mL of 0.800 M Co(NO₃)₂, which is equivalent to 0.0164 mol, and 21.0 mL of 0.800 M NaOH, which is equivalent to 0.0168 mol. Since the stoichiometric ratio between Co(NO₃)₂ and Co(OH)₂ is 1:1, the limiting reactant is Co(NO₃)₂. This means that 0.0164 mol of Co(OH)₂ will be formed.
To convert moles to grams, we need to use the molar mass of Co(OH)₂. The molar mass of Co(OH)₂ is 92.953 g/mol. Using the formula moles = mass / molar mass, we can rearrange the formula to mass = moles * molar mass. Plugging in the values, we have 0.0164 mol * 92.953 g/mol = 1.525 g. Therefore, approximately 1.525 grams of precipitate will be formed