Answer:
The answer to your question is 8.28 g of glucose
Explanation:
Data
Glucose (C₆H₁₂O₆) = ?
Ethanol (CH₃CH₂OH)
Carbon dioxide (CO₂) = 2.25 l
Pressure = 1 atm
T = 295°K
Reaction
C₆H₁₂O₆ ⇒ 2C₂H₅OH(l) +2CO₂(g)
- Calculate the number of moles
PV = nRT
Solve for n

Substitution

Simplification
n = 0.092
- Calculate the mass of glucose
1 mol of glucose --------------- 2 moles of carbon dioxide
x --------------- 0.092 moles
x = (0.092 x 1) / 2
x = 0.046 moles of glucose
Molecular weight of glucose = 180 g
180 g of glucose --------------- 1 mol
x g ---------------0.046 moles
x = (0.046 x 180) / 1
x = 8.28 g of glucose
Answer:
2Na2O2+2H2O⟶O2+4NaOH
2×78g 32g
156g of Na2O2 produces 32g of O2,
12g of Na2O2 produces =15632×12=10.66g.
Density of O2 at NTP=1.428g/mL.
DensityMass=Vol.
1.42810.66=7.46mL
Vol. of O2 at NTP is 7.46mL.
Explanation:
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molecules collide more frequently.
Chemical reaction mechanisms are based in the collision of molecules with certaing level of energy. More collisions implies grater probablity of reaction.