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1. Percentage yield is 64.75%

2. Atom economy is 48.39%

Explanation:

1. Determination of the percentage yield

Experimental yield = 24.8 g

Theoretical yield = 38.3 g

Percentage yield =?

Percentage yield = Experimental yield /Theoretical yield × 100

Percentage yield = 24.8 / 38.3 × Percentage yield = 2480 / 38.3

Percentage yield = 64.75%

2. Determination of the atom economy.

The balanced equation for the reaction is given below:

6CO₂ + 6H₂O —> C₆H₁₂O₆ + 6O₂

Next, we shall determine the mass of CO₂, H₂O and C₆H₁₂O₆ from the balanced equation.

Molar mass of CO₂ = 12 + (16×2)

= 12 + 32

= 44 g/mol

Mass of CO₂ from the balanced equation = 6 × 44 = 264 g

Molar mass of H₂O = (2×1) + 16

= 2 + 16

= 18 g/mol

Mass of H₂O from balanced equation = 6 × 18 = 108 g

Molar mass of C₆H₁₂O₆ = (6×12) + (12×1) + (6×16)

= 72 + 12 + 96

= 180 g/mol

Mass of C₆H₁₂O₆ from the balanced equation = 1 × 180 = 180 g

Next, we shall determine the total mass of the reactants. This is illustrated below:

Mass of CO₂ = 264 g

Mass of H₂O = 108 g

Total mass of reactants =?

Total mass of reactant = Mass of CO₂ + Mass of H₂O

Total mass of reactant = 264 + 108

Total mass of reactant = 372 g

Finally, we shall determine the atom economy. This can be obtained as follow:

Mass of desire atom (C₆H₁₂O₆) = 180 g

Total mass of reactants = 372 g

Atom economy =.?

Atom economy = mass of desire atom / total mass of reactant × 100

Atom economy = 180/372 × 100

Atom economy = 48.39%

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in the analysis of systems that involve fluid flow, we frequently encounter the combination of properties u and pv.