If a plant uses 44 grams of carbon dioxide and 18 grams of water to produce 32 grams of oxygen, how many grams of glucose are produced for the plant to use 6h2o + 6co2 --> c6h12o6 + 6o2

The balanced equation for the reaction is as follows 
6H₂O + 6CO₂ ---> C₆H₁₂O₆ + 6O₂

number of moles of CO₂ used - 44 g / 44 g/mol = 1 mol
number of moles of H₂O used - 18 g / 18 g/mol = 1 mol
stoichiometry of CO₂ to H₂O is 6:6 = 1:1
1 mol of CO₂ present and 1 mol of H₂O present therefore they are both fully used up in the reaction of molar ratio 1:1
number of moles of O₂ formed - 32 g/ 32 g/mol = 1 mol

stoichiometry of CO₂:H₂O:C₆H₁₂O₆:O₂ is 6:6:1:6
1 mol of CO₂ has reacted with 1 mol of H₂O to form 1 mol of O₂ and x mol of C₆H₁₂O₆

the number of C₆H₁₂O₆ moles is 1/6th of CO₂ moles used up
since CO₂ moles - 1 mol 
therefore C₆H₁₂O₆ moles formed - 1/6 mol = 0.167 mol
mass of glucose formed - 0.167 g x 180 g/mol = 30 g 
therefore 30 g of glucose is formed 

The density would be 1.622