12.00 moles of naclo3 will produce how many grams of O2? (Moles to Grams)
1 answer:
2NaClO₃ → 2NaCl + 3O₂
mole ratio of NaClO₃ to O₂ is 2 : 3
∴ if moles of NaClO₃ = 12 mol
then moles of O₂ =
= 18 mol
Mass of O₂ = mol of O₂ × molar mass of O₂
= 18 mol × 16 g/mol
= 288 g
So I wasn't sure which equation to use since you did not specify so I just used the decomposition reaction. If you should have used another reaction then just follow the same steps and you'll get your answer.
mole ratio of NaClO₃ to O₂ is 2 : 3
∴ if moles of NaClO₃ = 12 mol
then moles of O₂ =
= 18 mol
Mass of O₂ = mol of O₂ × molar mass of O₂
= 18 mol × 16 g/mol
= 288 g
So I wasn't sure which equation to use since you did not specify so I just used the decomposition reaction. If you should have used another reaction then just follow the same steps and you'll get your answer.
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The enthalpy of vaporization of Bromine is 15.4 kJ/mol. -7.7 kJ is the energy change when 80.2 g of Br₂ condenses to a liquid at 59.5°C.
<h3>What is Enthalpy of Vaporization ?</h3>The amount of enthalpy or energy that must be added to a liquid substance into gas substance is called Enthalpy of Vaporization. It is also known as Latent heat of vaporization.
<h3>How to find the energy change from enthalpy of vaporization ?</h3>To calculate the energy use this expression:
where,
Q = Energy change
n = number of moles
= Molar enthalpy of vaporization
Now find the number of moles
Number of moles (n) =
=
= 0.5 mol
Now put the values in above formula we get
[Negative sign is used because Br₂ condensed here]
= - (0.5 mol × 15.4 kJ/mol)
= - 7.7 kJ
Thus from the above conclusion we can say that The enthalpy of vaporization of Bromine is 15.4 kJ/mol. -7.7 kJ is the energy change when 80.2 g of Br₂ condenses to a liquid at 59.5°C.
Learn more about the Enthalpy of Vaporization here: brainly.com/question/13776849
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