Answer:
Mass of Glucose = 6.82 g
Explanation:
The balance chemical equation for given reaction is as;
6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂
To solve this reaction one should first find the limiting reagent. To do so e will calculate the moles of each reactant as,
Moles of CO₂ = Given Mass / M.Mass of CO₂
Moles of CO₂ = 10.0 g / 44.0 g/mol
Moles of CO₂ = 0.2272 moles
Similarly for H₂O,
Moles of H₂O = Given Mass / M.Mass of H₂O
Moles of H₂O = 5.0 g / 18.01 g/mol
Moles of H₂O = 0.2776 moles
Now, according to equation,
6 mole of CO₂ reacts with = 6 moles of H₂O
So,
0.2272 moles of CO₂ will react with = X moles of H₂O
Solving for X,
X = 0.2272 moles × 6 moles / 6 mole
X = 0.2272 moles of H₂O
As calculated above, we are provided with 0.2776 moles of H₂O while, we require only 0.2272 moles of it so it means that H₂O is in excess and CO₂ is the limiting reagent hence, CO₂ will control the yield of Glucose.
So,
The amount of Glucose produced is calculated by first finding its moles as,
According to equation,
6 mole of CO₂ produced = 1 mole of Glucose
So,
0.2272 moles of CaSO₄ will produce = X moles of Glucose
Solving for X,
X = 0.2272 moles × 1 mole / 6 mole
X = 0.0378 moles of Glucose
Now convert moles of Glucose to mass as,
Mass of Glucose = Moles × M.Mass
Mass of Glucose = 0.0378 × 180.15 g/mol
Mass of Glucose = 6.82 g
