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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.
Answer:- Formula of the hydrate is and it's name is Iron(III)sulfate pentahydrate.
Solution:- As per the given information, there is 18.4% water in the hydrate. If we assume the mass of the hydrate as 100 grams then there would be 18.4 grams of water and 81.6 grams of Iron(III)sulfate present in the hydrate.
Molar mass for Iron(III)sulfate is 399.88 gram per mol and the molar mass for water is 18.02 gram per mol.
We will calculate the moles of Iron(III)sulfate and water present in the compound on dividing their grams by their molar masses as:
=
=
Now, the next step is to calculate the mol ratio and for this we divide the moles of each by the least one of them means whose moles are less. Here, the moles of Iron(III)sulfate are less than moles of water. So, we divide the moles of each by 0.204.
= 1
= 5
There is 1:5 mol ratio between Iron(III)sulfate and water. So, the formula of the hydrate is and the name of the hydrate is Iron(III)sulfate pentahydrate.