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
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Answer:
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
Thus, in terms of pressures, the rate becomes:
Thus, the rate of change for the partial pressure of ammonia turns out:
![r_{NH_3}=2*(-r_{N_2H_4})\\r_{NH_3}=2*[-(-70torr/h)]\\r_{NH_3}=140torr/h](https://tex.z-dn.net/?f=r_%7BNH_3%7D%3D2%2A%28-r_%7BN_2H_4%7D%29%5C%5Cr_%7BNH_3%7D%3D2%2A%5B-%28-70torr%2Fh%29%5D%5C%5Cr_%7BNH_3%7D%3D140torr%2Fh)
The rate of decrease of partial pressure of urea is taken negative as it is a reactant whereas ammonia a product which has 2 as its stoichiometric coefficient.
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