Movement Across a Membrane and Energy. There are two major ways that molecules can be moved across a membrane, and the distinction has to do with whether or not cell energy is used. Passive mechanisms like diffusion use no energy, while active transport requires energy to get done.
2C6H14 + 13O2 ---> 6CO2 +14H2O
M(C6H14)=12.011*6 +1.008*14 ≈ 86.17 g/mol
86.17 g C6H14 is 1 mole.
2C6H14 + 13O2 ---> 6CO2 +14H2O
from reaction 2 mol 6 mol
from the problem 1 mol 3 mol
M(CO2)= 12.011 + 2*15.999= 44.009 g/mol
3 mol CO2*44.009 g/1 mol CO2 ≈ 132.0 g CO2
Answer : 132.0 g CO2
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Answer:
3.2 g O₂
Explanation:
To find the mass of O₂, you need to (1) convert grams H₂O to moles H₂O (via molar mass), then (2) convert moles H₂O to moles O₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles O₂ to grams O₂ (via molar mass). It is important to arrange the ratios/conversions in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 2 sig figs to reflect the sig figs of the given value (3.6 g).
Molar Mass (H₂O): 2(1.008 g/mol) + 15.998 g/mol
Molar Mass (H₂O): 18.014 g/mol
2 H₂O -----> 2 H₂ + 1 O₂
Molar Mass (O₂): 2(15.998 g/mol)
Molar Mass (O₂): 31.996 g/mol
3.6 g H₂O 1 mole 1 mole O₂ 31.996 g
---------------- x --------------- x --------------------- x --------------- = 3.2 g O₂
18.014 g 2 moles H₂O 1 mole