Step 1
The reaction involved:
6 CO2 + 6 H2O ⇨ C6H12O6 + 6 O2 (completed and balanced)
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Step 2
Information provided:
88.0 g CO2
64.0 g water (H2O)
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Information needed:
The molar masses of:
CO2) 44.0 g/mol
H2O) 18.0 g/mol
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Step 3
Procedure:
By stoichiometry,
6 CO2 + 6 H2O ⇨ C6H12O6 + 6 O2
6 x 44.0 g CO2 ---------------------- 6 x 18.0 g H2O
88.0 g CO2 ---------------------- X
X = 88.0 g CO2 x 6 x 18.0 g H2O/6 x 44.0 g CO2
X = 36 g H2O
For 88.0 g of CO2, 36 g of H2O is needed, but there is 64.0 g of water.
Therefore, the excess reactant is H2O and the limiting reactant is CO2.
Answer:
The limiting reactant = CO2
The excess reactant = H2O