If 30.00 g of calcium acetate reacts with an excess of aluminum nitrate, how
many grams of each product would be produced? When this was performed
in a lab setting, we were able to produce an actual yield of 29.45 g of
calcium nitrate. What would be the percent yield of this reaction?

Expected mass of each product from the reaction

Calcium Nitrate [Ca(NO₃)₂] = 31.11 g

Aluminium Acetate [Al(CH₃COO)₃] = 25.80 g

Percent yield = 94.66%

Explanation:

The balanced chemical reaction for when calcium acetate reacts with Aluminium nitrate is given as

3Ca(CH₃COO)₂ + 2Al(NO₃)₃ → 3Ca(NO₃)₂ + 2Al(CH₃COO)₃

30.00 g of Calcium Acetate reacts with excess Aluminium nitrate. This means that the calcium acetate is in short supply and is the limiting reagent. The limiting reagent is the reactant that is used up in the process of the reaction. The limiting reagent determines how much of other reactants will react and how much products will be formed.

Converting the 30.00 g of Calcium Acetate into number of moles.

Number of moles = (Mass)/(Molar Mass)

Molar mass of Calcium Acetate = 158.17 g/mol

Number of moles of Calcium Acetate = (30/158.17) = 0.1897 moles

To now find the number of moles of each product formed, from the stoichiometric balance of the reaction,

3Ca(CH₃COO)₂ + 2Al(NO₃)₃ → 3Ca(NO₃)₂ + 2Al(CH₃COO)₃

3 moles of Calcium Acetate gives 3 moles of Calcium Nitrate [Ca(NO₃)₂]

0.1897 mole of Calcium Acetate will also give 0.1897 mole of Calcium Nitrate.

3 moles of Calcium Acetate gives 2 moles of Aluminium Acetate [Al(CH₃COO)₃]

0.1897 mole of Calcium Acetate will give (0.1897×2/3) mole of Aluminium Acetate; That is, 0.1264 mole of Aluminium Acetate.

To now find the mass of each product produced,

Mass = (Number of moles) × (Molar Mass)

Number of moles of Calcium Nitrate produced = 0.1897 mole

Molar mass of Calcium Nitrate = 164.088 g/mol

Mass of Calcium Nitrate produced

= 0.1897 × 164.088 = 31.13 g

Number of moles of Aluminium Acetate produced = 0.1264 mole

Molar mass of Aluminium Acetate = 204.11 g/mol

Mass of Aluminium Acetate produced

= 0.1264 × 204.11 = 25.80 g

b) Percent yield

= 100% × (Actual yield)/(Theoretical yield)

Actual yield of Calcium Nitrate = 29.45 g

Theoretical yield of Calcium Nitrate = 31.11 g

Percent yield = 100% × (29.45/31.11) = 94.66%

Hope this Helps!!!