Respuesta :
Answer:
The mass in grams of glucose produced when 132.0 g of CO2 reacts with an excess of water is 90.1 grams
Explanation:
The chemical equation for the reaction is
6H₂O + 6CO₂ → C₆H₁₂O₆ + 6O₂
From the reaction, it is seen that 6 moles of H₂O reacts ith 6 moles of CO₂ to produce 1 mole of glucose C₆H₁₂O₆ and 6 moles oxygen gas
The molar mass of CO₂ = 44.01 g/mol
There fpre 132.0 g contains 132.0/44.01 moles or ≅ 3 moles
However since 6 moles of CO₂ produces 1 mole of O₂, then 3 moles of CO₂ will prduce 1/6×3 or 0.5 moles of C₆H₁₂O₆
and since the molar mass (or the mass of one mole) of C₆H₁₂O₆ is 180.2 grams/mole then 0.5 mole of C₆H₁₂O₆ will have a mass of
mass of 1 mole C₆H₁₂O₆ = 180.2 g
mass of 0.5 mole C₆H₁₂O₆ = 180.2 g × 0.5 = 90.1 grams
Mass of glucose produced = 90.1 grams
The mass of glucose produced when 132.0 g of CO₂ reacts with an excess of water is 90.1 grams.
Balanced Chemical Equation:
6H₂O + 6CO₂ → C₆H₁₂O₆ + 6O₂
⇒ 6 moles of H₂O reacts with 6 moles of CO₂ to produce 1 mole of glucose C₆H₁₂O₆ and 6 moles oxygen gas.
The molar mass of CO₂ = 44.01 g/mol
Thus, 132.0 g contains [tex]\frac{132.0}{44.01} =3 \text{ moles}[/tex]
- 6 moles of CO₂ produces 1 mole of O₂.
Then, 3 moles of CO₂ will produce [tex]1/6 *3 = 0.5 \text{ moles}[/tex] of C₆H₁₂O₆
The molar mass of C₆H₁₂O₆ is 180.2 g/mol.
- 0.5 mole of C₆H₁₂O₆ will have a mass of
Mass of 0.5 mole C₆H₁₂O₆ = [tex]180.2 g * 0.5 = 90.1 \text{ grams}[/tex]
Mass of glucose produced = 90.1 grams
Find more information about Molar mass here:
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