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iren2701 [21]
3 years ago
9

What cycloalkanes are present in a steroid ?

Chemistry
1 answer:
11Alexandr11 [23.1K]3 years ago
3 0
Steroids are biologically very important compounds. They are made up of more that seventeen carbons. The structure of Steroid is comprised of a fused cycloalkane ring system. The simplest steroid and the main skeleton of all steroids is Gonane.

Gonane is made up of four cycloalkane rings i.e. tetracyclic hydrocarbon ring structure. Among these four rings 3 are cyclohexanes (designated as A, B and C) and one is cyclopentane (Designated as D). These four rings fuse to form a <span>cyclopentaperhydrophenanthrene.

The Nucleus (Gonane) of steroids along with some derivatives are shown below,</span>

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When 125 grams of FeO react with 25.0 grams of Al, how many grams of Fe can be produced? FeO + Al → Fe + Al2O3 25.9 g Fe 38.7 g
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<u>Answer:</u> The mass of iron produced will be 77.6 grams

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\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

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Given mass of FeO = 125 g

Molar mass of FeO = 71.8 g/mol

Putting values in equation 1, we get:

\text{Moles of FeO}=\frac{125g}{71.8g/mol}=1.74mol

  • <u>For aluminium:</u>

Given mass of aluminium = 25.0 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

\text{Moles of aluminium}=\frac{25.0g}{27g/mol}=0.93mol

The given chemical reaction follows:

3FeO+2Al\rightarrow 3Fe+Al_2O_3

By Stoichiometry of the reaction:

2 moles of aluminium metal reacts with 3 mole of FeO

So, 0.93 moles of aluminium metal will react with = \frac{3}{2}\times 0.93=1.395mol of FeO

As, given amount of FeO is more than the required amount. So, it is considered as an excess reagent.

Thus, aluminium metal is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of aluminium metal produces 3 mole of iron metal

So, 0.93 moles of aluminium metal will produce = \frac{3}{2}\times 0.93=1.395moles of iron metal

  • Now, calculating the mass of iron metal from equation 1, we get:

Molar mass of iron = 55.85 g/mol

Moles of iron = 1.395 moles

Putting values in equation 1, we get:

1.395mol=\frac{\text{Mass of iron}}{55.85g/mol}\\\\\text{Mass of iron}=(1.395mol\times 55.85g/mol)=77.6g

Hence, the mass of iron produced will be 77.6 grams

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