PLEASE HELP WITH THESE 2 QUESTIONS

4. Precipitation of Calcium Carbonate:

Calcium chloride (CaCl,) and sodium carbonate (Na,CO₂) react in solution to form equation is:CaCl₂ (aq) + Na₂CO₂ (aq) → CaCO₂ (s) + 2NaCl (aq) calcium carbonate (CaCO₂) and sodium chloride (NaCl). The balanced chemical A chemist mixes 100.0 mL of a 0.100 M CaCI, solution with 100.0 mL of a 0.200 M Na₂CO solution. After filtration and drying, the chemist recovers 0.650 grams of calcium carbonate. What is the percent yield of the precipitation reaction?

5. Production of Ethanol: A biochemist ferments glucose (CHO) to produce ethanol (C₂H₂OH) and carbon dioxide (CO₂) in a fermentation process. The balanced chemical equation is: CH₁₂O (aq) → 2C₂H₂OH(aq) + 2CO₂ (g) A large-scale fermentation produces 10,000 liters of a 10% (v/v) ethanol solution. Assuming the density of the solution is 0.79 kg/L, what is the percent yield of the fermentation reaction if the theoretical yield is 18,420 kg of ethanol?

4. To find the percent yield of the precipitation reaction, we need to compare the actual yield of calcium carbonate obtained with the theoretical yield we would expect based on the balanced chemical equation.

First, let's calculate the moles of calcium carbonate obtained:

Molar mass of CaCO₂ = 40.08 g/mol + 12.01 g/mol + (16.00 g/mol × 2) = 100.08 g/mol

Number of moles of CaCO₂ = Mass / Molar mass = 0.650 g / 100.08 g/mol

Next, let's calculate the moles of calcium chloride (CaCl₂) and sodium carbonate (Na₂CO₃) used in the reaction:

Moles of CaCl₂ = Volume × Concentration = 100.0 mL × 0.100 mol/L / 1000 mL/L = 0.0100 mol

Moles of Na₂CO₃ = Volume × Concentration = 100.0 mL × 0.200 mol/L / 1000 mL/L = 0.0200 mol

Since the stoichiometry of the balanced equation is 1:1 for CaCl₂ and Na₂CO₃ to CaCO₂, the limiting reactant is CaCl₂.

Theoretical yield of CaCO₂ = Moles of limiting reactant × molar ratio of CaCO₂ to limiting reactant × molar mass of CaCO₂

= 0.0100 mol CaCl₂ × (1 mol CaCO₂ / 1 mol CaCl₂) × 100.08 g/mol

= 1.0008 g

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

= (0.650 g / 1.0008 g) × 100

= 65.00%

Therefore, the percent yield of the precipitation reaction is 65.00%.

5. To find the percent yield of the fermentation reaction, we need to compare the actual yield of ethanol obtained with the theoretical yield we would expect based on the balanced chemical equation.

First, let's calculate the mass of ethanol obtained in the fermentation:

Mass of ethanol = Volume × Density = 10,000 L × 0.79 kg/L = 7,900 kg

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

= (7,900 kg / 18,420 kg) × 100

= 42.86%

Therefore, the percent yield of the fermentation reaction is 42.86%.