A student boils 9.43 grams of an unknown substance into vapor. If completely vaporizing the substance required 8,720 joules of e
2 answers:
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<u>Answer:</u> The theoretical yield of the lithium chlorate is 1054.67 grams
<u>Explanation:</u>
To calculate the mass for given number of moles, we use the equation:
Actual moles of lithium chlorate = 9.45 moles
Molar mass of lithium chlorate = 90.4 g/mol
Putting values in above equation, we get:
To calculate the theoretical yield of lithium chlorate, we use the equation:
Actual yield of lithium chlorate = 854.28 g
Percentage yield of lithium chlorate = 81.0 %
Putting values in above equation, we get:
Hence, the theoretical yield of the lithium chlorate is 1054.67 grams
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.